Winter holiday camp 1.19
Article Directory
A - Maya Calendar
During his last sabbatical, professor M. A. Ya made a surprising discovery about the old Maya calendar. From an old knotted message, professor discovered that the Maya civilization used a 365 day long year, called Haab, which had 19 months. Each of the first 18 months was 20 days long, and the names of the months were pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu. Instead of having names, the days of the months were denoted by numbers starting from 0 to 19. The last month of Haab was called uayet and had 5 days denoted by numbers 0, 1, 2, 3, 4. The Maya believed that this month was unlucky, the court of justice was not in session, the trade stopped, people did not even sweep the floor.
For religious purposes, the Maya used another calendar in which the year was called Tzolkin (holly year). The year was divided into thirteen periods, each 20 days long. Each day was denoted by a pair consisting of a number and the name of the day. They used 20 names: imix, ik, akbal, kan, chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban, eznab, canac, ahau and 13 numbers; both in cycles.
Notice that each day has an unambiguous description. For example, at the beginning of the year the days were described as follows:
1 imix, 2 ik, 3 akbal, 4 kan, 5 chicchan, 6 cimi, 7 manik, 8 lamat, 9 muluk, 10 ok, 11 chuen, 12 eb, 13 ben, 1 ix, 2 mem, 3 cib, 4 caban, 5 eznab, 6 canac, 7 ahau, and again in the next period 8 imix, 9 ik, 10 akbal . . .
Years (both Haab and Tzolkin) were denoted by numbers 0, 1, : : : , where the number 0 was the beginning of the world. Thus, the first day was:
Haab: 0. pop 0
Tzolkin: 1 imix 0
Help professor M. A. Ya and write a program for him to convert the dates from the Haab calendar to the Tzolkin calendar.
Input
The date in Haab is given in the following format:
NumberOfTheDay. Month Year
The first line of the input file contains the number of the input dates in the file. The next n lines contain n dates in the Haab calendar format, each in separate line. The year is smaller then 5000.
Output
The date in Tzolkin should be in the following format:
Number NameOfTheDay Year
The first line of the output file contains the number of the output dates. In the next n lines, there are dates in the Tzolkin calendar format, in the order corresponding to the input dates.
Sample Input
3
10. zac 0
0. pop 0
10. zac 1995
Sample Output
3
3 chuen 0
1 imix 0
9 cimi 2801
$\mathbf{topic outline} $
The Maya used the Haab calendar. There are 365 days in a year and 19 months in a year. The first eighteen months have 20 days per month. The last month of the Haab calendar is called uayet. This month is only 5 days. For religious reasons, the Mayans still Using the Tzolkin calendar, the year is divided into 13 different periods, each with 20 days. 0 means the beginning of the world, so the first day is expressed as Haab: 0. pop 0; Tzolkin: 1 imix 0, write a program to convert Haab calendar into Tzolkin calendar.
Idea : Because you want to convert the calendar year, you need to calculate how many days have passed since the beginning of the haab calendar from the world to the given date. The storage month is stored in a string array.
Main code
for(int j = 0; j<19;++j){
if(p[i].month==haab[j]){
t = j;
break;
}
}
cnt = p[i].year*365+t*20+p[i].day;//计数共过了多少天
year = cnt/260;// 按tzolki算经历了多少年
cnt = cnt%260;//算出剩余总天数
day =cnt%13+1;
month = cnt%20;//月份的小标（在字符串的位置）
Complete code
#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
const int N = 5000;
struct data{
int year,day;
string month;
}p[N];
int n;
int main(){
string haab[] = {"pop","no","zip","zotz","tzec",
"xul","yoxkin","mol","chen","yax","zac","ceh",
"mac","kankin","muan","pax","koyab","cumhu","uayet"};
string tzo[] = {"imix","ik","akbal","kan","chicchan","cimi",
"manik","lamat","muluk","ok","chuen","eb","ben","ix",
"mem","cib","caban","eznab","canac","ahau"};
cin>>n;
cout<<n<<endl;
for(int i = 1; i <= n;++i){
scanf("%d. ",&p[i].day);
cin>>p[i].month;
cin>>p[i].year;
int year,month,day,t,cnt;
for(int j = 0; j<19;++j){
if(p[i].month==haab[j]){
t = j;
break;
}
}
cnt = p[i].year*365+t*20+p[i].day;//计数共过了多少天
year = cnt/260;// 按tzolki算经历了多少年
cnt = cnt%260;//算出剩余总天数
day =cnt%13+1;
month = cnt%20;
cout<<day<<" "<<tzo[month]<<" "<<year<<endl;
}
return 0;
}
B - Diplomatic License
In an effort to minimize the expenses for foreign affairs the countries of the world have argued as follows. It is not enough that each country maintains diplomatic relations with at most one other country, for then, since there are more than two countries in the world, some countries cannot communicate with each other through (a chain of) diplomats.
Now, let us assume that each country maintains diplomatic relations with at most two other countries. It is an unwritten diplomatic “must be” issue that every country is treated in an equal fashion. It follows that each country maintains diplomatic relations with exactly two other countries.
International topologists have proposed a structure that fits these needs. They will arrange the countries to form a circle and let each country have diplomatic relations with its left and right neighbours. In the real world, the Foreign Office is located in every country’s capital. For simplicity, let us assume that its location is given as a point in a two-dimensional plane. If you connect the Foreign Offices of the diplomatically related countries by a straight line, the result is a polygon.
It is now necessary to establish locations for bilateral diplomatic meetings. Again, for diplomatic reasons, it is necessary that both diplomats will have to travel equal distances to the location. For efficiency reasons, the travel distance should be minimized. Get ready for your task!
Input
The input contains several testcases. Each starts with the number n of countries involved. You may assume that n>=3 is an odd number. Then follow n pairs of x- and y-coordinates denoting the locations of the Foreign Offices. The coordinates of the Foreign Offices are integer numbers whose absolute value is less than 1012. The countries are arranged in the same order as they appear in the input. Additionally, the first country is a neighbour of the last country in the list.
Output
For each test case output the number of meeting locations (=n) followed by the x- and y-coordinates of the locations. The order of the meeting locations should be the same as specified by the input order. Start with the meeting locations for the first two countries up to the last two countries. Finally output the meeting location for the n-th and the first country.
Sample Input
5 10 2 18 2 22 6 14 18 10 18
3 -4 6 -2 4 -2 6
3 -8 12 4 8 6 12
Sample Output
5 14.000000 2.000000 20.000000 4.000000 18.000000 12.000000 12.000000 18.000000 10.000000 10.000000
3 -3.000000 5.000000 -2.000000 5.000000 -3.000000 6.000000
3 -2.000000 10.000000 5.000000 10.000000 -1.000000 12.000000
Hint
Note that the output can be interpreted as a polygon as well. The relationship between the sample input and output polygons is illustrated in the figure on the page of Problem 1940. To generate further sample input you may use your solution to that problem.
The main idea of the topic: Given n (odd number) countries, n pairs of coordinates represent the position of the Ministry of Foreign Affairs, and the x and y coordinates of the location of bilateral diplomatic conferences between countries are given. The order of the output of the meeting place should be the order given by the input the same. From the meeting place of the top two countries to the meeting place of the last two countries, the meeting place of the nth country and the first country are output at the end
Idea: The title suggests that a circle formed by the national quality inspection can be regarded as a polygon, and the Ministry of Foreign Affairs must make the distance between two neighboring countries and the Ministry of Foreign Affairs equal, so we only need to ask for the midpoint coordinates of each pair of neighboring countries. can
Complete code:
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n;
const int N = 1e4+6;
std::pair<double,double>loc[N];
int main(){
while(cin>>n){
cout<<n;
for(int i = 1;i<=n;++i){
cin>>loc[i].first>>loc[i].second;
}
for(int i = 1;i<=n-1;++i){
printf(" %.6f %.6f",(loc[i].first+loc[i+1].first)/2,(loc[i].second+loc[i+1].second)/2);
}
printf(" %.6f %.6f",(loc[1].first+loc[n].first)/2,(loc[1].second+loc[n].second)/2);
cout<<endl;
}
return 0;
}
C - “Accordian” Patience
You are to simulate the playing of games of ``Accordian’’ patience, the rules for which are as follows:
Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate neighbour on the left, or matches the third card to the left, it may be moved onto that card. Cards match if they are of the same suit or same rank. After making a move, look to see if it has made additional moves possible. Only the top card of each pile may be moved at any given time. Gaps between piles should be closed up as soon as they appear by moving all piles on the right of the gap one position to the left. Deal out the whole pack, combining cards towards the left whenever possible. The game is won if the pack is reduced to a single pile.
Situations can arise where more than one play is possible. Where two cards may be moved, you should adopt the strategy of always moving the leftmost card possible. Where a card may be moved either one position to the left or three positions to the left, move it three positions.
Input
Input data to the program specifies the order in which cards are dealt from the pack. The input contains pairs of lines, each line containing 26 cards separated by single space characters. The final line of the input file contains a # as its first character. Cards are represented as a two character code. The first character is the face-value (A=Ace, 2-9, T=10, J=Jack, Q=Queen, K=King) and the second character is the suit (C=Clubs, D=Diamonds, H=Hearts, S=Spades).
Output
One line of output must be produced for each pair of lines (that between them describe a pack of 52 cards) in the input. Each line of output shows the number of cards in each of the piles remaining after playing ``Accordian patience’’ with the pack of cards as described by the corresponding pairs of input lines.
Sample Input
QD AD 8H 5S 3H 5H TC 4D JH KS 6H 8S JS AC AS 8D 2H QS TS 3S AH 4H TH TD 3C 6S
8C 7D 4C 4S 7S 9H 7C 5D 2S KD 2D QH JD 6D 9D JC 2C KH 3D QC 6C 9S KC 7H 9C 5C
AC 2C 3C 4C 5C 6C 7C 8C 9C TC JC QC KC AD 2D 3D 4D 5D 6D 7D 8D TD 9D JD QD KD
AH 2H 3H 4H 5H 6H 7H 8H 9H KH 6S QH TH AS 2S 3S 4S 5S JH 7S 8S 9S TS JS QS KS
#
Sample Output
6 piles remaining: 40 8 1 1 1 1
1 piles remaining: 52
The main idea: simulate a card game, the rules are as follows
Please simulate the "Accordian" Patience game, the rules are as follows:
• Players will deal a deck of cards one by one, in a row from left to right, without overlapping. As long as a playing card matches the first card from the left or the third card from the left, move this playing card on top of the matched card. The so-called two-card match is the same value (number or letter) or the same suit of the two cards. Whenever you move a card, check again to see if the card can continue to move to the left. You can only move the card at the top of the deck each time. In this game, two piles can be turned into one pile. If according to the rules, the cards from the right pile can be moved to the left pile one by one, and it can become one pile. The game moves the cards to the left as much as possible. If there is only one deck at the end, the player wins.
• During the game, players may encounter a situation where there are multiple choices at once. When both cards can be moved, the leftmost card is moved. If a card can move one position to the left or three positions to the left, move it three positions.
Idea: Linked list problem, use pre[] to point to the previous card, last[] to point to the next card, s [] [] [] The third dimension stores the face value and suit of the card, simulating the game process, and did not make it by yourself. I have some lack of mastery of the code implementation of the linked list. After reading Le Ge's code, it is too strong. .
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 60;
char s[N][N][N];
int pre[N],last[N],mp[N],sum[N],ans[N];
int getp3(int x){
int p1 = pre[x],p2 = pre[p1],p3 = pre[p2];
return p3;
}
int getp1(int x){
int p1 = pre[x];
return p1;
}
int main(){
while(1){
scanf("%s",s[1][1]);
if(s[1][1][0]=='#') break;
last[1] = 2,sum[1]=1,last[0]=1;
for(int i = 2;i<=52;++i){
scanf("%s",s[i][1]);
pre[i] = i-1;
last[i] = i+1;
sum[i] = 1;
}
for(int i = 2;i<=52;i = last[i]){
int p = getp3(i);
if(p>0&&(s[p][sum[p]][0]==s[i][sum[i]][0]||
s[p][sum[p]][1]==s[i][sum[i]][1])){
sum[p]++;
s[p][sum[p]][0]=s[i][sum[i]][0];
s[p][sum[p]][1]=s[i][sum[i]][1];
sum[i]--;
if(sum[i]==0){
last[pre[i]] = last[i];
pre[last[i]] = pre[i];
}
i = p-1;
continue;
}
p = getp1(i);
if(p>0&&(s[p][sum[p]][0]==s[i][sum[i]][0]||
s[p][sum[p]][1]==s[i][sum[i]][1])){
sum[p]++;
s[p][sum[p]][0]=s[i][sum[i]][0];
s[p][sum[p]][1]=s[i][sum[i]][1];
sum[i]--;
if(sum[i]==0){
last[pre[i]] = last[i];
pre[last[i]] = pre[i];
}
i = p-1;
continue;
}
}
int cnt = 0;
for(int i = 1;i<=52;++i){
if(sum[i]) ans[cnt++] = sum[i];
}
printf("%d piles remaining: %d",cnt,ans[0]);
for(int i = 1;i<cnt;++i){
printf(" %d",ans[i]);
}
printf("\n");
}
return 0;
}
D - Broken Keyboard (a.k.a. Beiju Text)
The main idea: Someone is typing a long text with a broken keyboard. The problem with this keyboard is that from time to time the "Home" key or "End" key will
Automatically press when text. You are not aware of this problem, because you only focus on the text and you don't even turn on the display. After you finish typing, you turn on the monitor and see the text on the screen. In Chinese, we call it tragedy. Please find the text of the tragedy
Idea: Use an array to simulate a linked list, the next array replaces the next pointer in the linked list, next[i] stores the next character subscript of the i-th character
next[i] = i+1, set the variable cur to represent the cursor, cur is not the position i currently traversed , it means that the character at position i should be inserted to the right of cur . If the current character is '[', the cursor cur jumps to the front of the string, ie cur =0; if']', the cursor jumps to the end of the string, ie cur=last , where the variable last saves the current character The rightmost subscript of the string.
The main code:
str="*"+str;
for(int i =1;str[i];++i){
if(str[i]=='[') cur = 0;//光标跳到字符串前
else if(str[i]==']') cur = last;//跳到字符串后
else{
Next[i] = Next[cur];//插入链表
Next[cur] = i;
if(cur==last) last = i;
cur = i;
}
}
Complete code:
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 1e5+6;
int Next[N];
int main(){
string str;
while(getline(cin,str)){
int cur = 0,last = 0;
Next[0]=0;
str="*"+str;
for(int i =1;str[i];++i){
if(str[i]=='[') cur = 0;//光标跳到字符串前
else if(str[i]==']') cur = last;//跳到字符串后
else{
Next[i] = Next[cur];//插入链表
Next[cur] = i;
if(cur==last) last = i;
cur = i;
}
}
for(int i = Next[0];i;i = Next[i]){
cout<<str[i];
}
cout<<endl;
}
return 0;
}
E - Satellites
The main idea of the topic: Give the surface distance between the artificial satellite and the earth, and the angle between the two artificial satellites and the center of the earth. The min string represents the angle in minutes, and deg is in degrees.
Idea: Use the formula to solve [External link image transfer failed, the source site may have an anti-leech link mechanism, it is recommended to save the image and upload it directly (img-OeYW0deH-1611305756156)(C:\Users\29252\Pictures\topic\E2.png )]
Pay attention to situations where the angle is greater than 180 degrees
Complete code:
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
double s,a;
string str;
double r = 6440;
int main(){
while(cin>>s>>a>>str){
if(str[0]=='m') a/=60;
double arc_dist = 2*PI*(r+s)*(a/360);
double chord_dist = (r+s)*sin(a*PI/360)*2;
if(a>180) arc_dist = 2*PI*(r+s)-arc_dist;
printf("%.6f %.6f\n",arc_dist,chord_dist);
}
return 0;
}
F - Fourth Point !!
The main idea of the topic: Given the ( x , y ) coordinates of the endpoints of two adjacent sides in the parallelogram , please find the ( x , y ) coordinates of the fourth point .
Idea: Given the endpoint coordinates of two adjacent sides in the parallelogram, find the coordinates of the fourth point. It should be noted that for the endpoint coordinates of two adjacent sides, the coordinates of two points will be duplicated; therefore, it is necessary to determine which two points are duplicated.
Suppose the endpoint coordinates of two adjacent sides in the given parallelogram are ( x 0, y 0), ( x 1, y 1), ( x 2, y 2) and ( x 3, y 3), ( x 0, y 0 )= ( x 3, y 3), find the coordinates of the fourth point ( x**a , y**b ), then x**a - x 2 = x 1- x 0, y **a - y 2= y 1- y 0; get x**a = x 2+ x 1- x 0, y**a= y2+y1-y0。
Complete code:
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
double x,y;
};
int main(){
struct node a,b,c,d;
int x,y;
while(~scanf("%lf%lf%lf%lf%",&a.x,&a.y,&b.x,&b.y)){
scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
if(a.x==c.x&&a.y==c.y)
swap(a,b);
if(a.x==d.x&&a.y==d.y)
swap(a,b),swap(c,d);
if(b.x==d.x&&b.y==d.y)
swap(c,d);
printf("%.3f %.3f\n",a.x+d.x-c.x,a.y+d.y-c.y);
}
return 0;
}
G - The Circumference of the Circle
To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don’t?
You are given the cartesian coordinates of three non-collinear points in the plane.
Your job is to calculate the circumference of the unique circle that intersects all three points.
Input
The input will contain one or more test cases. Each test case consists of one line containing six real numbers x1,y1, x2,y2,x3,y3, representing the coordinates of the three points. The diameter of the circle determined by the three points will never exceed a million. Input is terminated by end of file.
Output
For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.
Sample Input
0.0 -0.5 0.5 0.0 0.0 0.5
0.0 0.0 0.0 1.0 1.0 1.0
5.0 5.0 5.0 7.0 4.0 6.0
0.0 0.0 -1.0 7.0 7.0 7.0
50.0 50.0 50.0 70.0 40.0 60.0
0.0 0.0 10.0 0.0 20.0 1.0
0.0 -500000.0 500000.0 0.0 0.0 500000.0
Sample Output
3.14
4.44
6.28
31.42
62.83
632.24
3141592.65
The main idea of the topic: Given the Cartesian coordinates of three non-collinear points on the plane, calculate the perimeter of the only circle where these three points intersect
Idea: Let 3 points be ( x 0, y 0), ( x 1, y 1) and ( x 2, y 2 ), and the center of the circle is ( x**m , y**m ).
Formula Derivation:
Complete code:
#include<iostream>
#include<cmath>
using namespace std;
double ax,ay,bx,by,cx,cy,p;
double getdis(double x,double y,double xx,double yy){
return sqrt(pow(x-xx,2)+pow(y-yy,2));
}
int main(){
while(cin>>ax>>ay>>bx>>by>>cx>>cy){
double a = getdis(ax,ay,bx,by);
double b = getdis(bx,by,cx,cy);
double c = getdis(ax,ay,cx,cy);
p = (a+b+c)/2;
double s = sqrt(p*(p-a)*(p-b)*(p-c));
double d = a*b*c/2.0/s;
printf("%.2f\n",d*acos(-1.0));
}
}
H - Titanic
It is a historical fact that during the legendary voyage of “Titanic” the wireless telegraph machine had delivered 6 warnings about the danger of icebergs. Each of the telegraph messages described the point where an iceberg had been noticed. The first five warnings were transferred to the captain of the ship. The sixth one came late at night and a telegraph operator did not notice that the coordinates mentioned were very close to the current ship’s position.
Write a program that will warn the operator about the danger of icebergs!
Input
The input messages are of the following format:
Message #<n>.
Received at <HH>:<MM>:<SS>.
Current ship's coordinates are
<X1>^<X2>'<X3>" <NL/SL>
and <Y1>^<Y2>'<Y3>" <EL/WL>.
An iceberg was noticed at
<A1>^<A2>'<A3>" <NL/SL>
and <B1>^<B2>'<B3>" <EL/WL>.
===
Here is a positive integer, :: is the time of the message reception, ^’" <NL/SL> and ^’" <EL/WL> means “X1 degrees X2 minutes X3 seconds of North (South) latitude and Y1 degrees Y2 minutes Y3 seconds of East (West) longitude.”
Output
Your program should print to the output file message in the following format:
The distance to the iceberg: <s> miles.
Where should be the distance between the ship and the iceberg, (that is the length of the shortest path on the sphere between the ship and the iceberg). This distance should be printed up to (and correct to) two decimal digits. If this distance is less than (but not equal to!) 100 miles the program should print one more line with the text:
DANGER!
Sample Input
Message #513.
Received at 22:30:11.
Current ship's coordinates are
41^46'00" NL
and 50^14'00" WL.
An iceberg was noticed at
41^14'11" NL
and 51^09'00" WL.
===
Sample Output
The distance to the iceberg: 52.04 miles.
DANGER!
Hint
For simplicity of calculations assume that the Earth is an ideal sphere with the diameter of 6875 miles completely covered with water. Also you can be sure that lines in the input file break exactly as it is shown in the input samples. The ranges of the ship and the iceberg coordinates are the same as the usual range for geographical coordinates, i.e. from 0 to 90 degrees inclusively for NL/SL and from 0 to 180 degrees inclusively for EL/WL.
Main idea: This question requires you to calculate the distance between two points on a sphere. The formula for calculating the distance on the sphere is directly used. If the distance is less than 100 miles, output: "DANGER!"
Idea: Knowing the latitude and longitude of two points A and B on the sphere , which are ( wA , jA ) and ( wB , jB ) respectively, the formula for calculating the distance between A and B : dist ( A , B ) =
*R**arccos(cos( wA )*cos( wB ) cos( jA - jB )+sin( wA ) sin( wB )); where R is the radius of the sphere, and the default'N' and'E ' are positive directions ,'S' and'W' are negative directions.
The input processing is relatively troublesome. First convert the degrees, minutes, and seconds of the latitude and longitude into degrees, and then take the sign of the east-west longitude and the north-south latitude. When calculating the distance, degrees are converted to radians; then the distance between the ship and the iceberg is calculated according to the formula for the distance between two points on the ball.
The main code:
double solve(double a,double b,double c,char s[])
{
a=a+c/3600+b/60;
if(s[2]=='S' || s[2]=='W')
a=-a;
a=a*pi/180;
return a;
}
Complete code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
char s[1000],s1[1000];
const double pi=acos(-1.0);
double solve(double a,double b,double c,char s[])
{
a=a+c/3600+b/60;
if(s[2]=='S' || s[2]=='W')
a=-a;
a=a*pi/180;
return a;
}
int dblcmp(double k)
{
if(fabs(k)<eps)
return 0;
return k>0?1:-1;
}
int main()
{
double a,b,c;
double a1,b1,a2,b2;
while(gets(s))
{
gets(s);
gets(s);
scanf("%lf^%lf'%lf",&a,&b,&c);
gets(s);
a1=solve(a,b,c,s);
scanf("and %lf^%lf'%lf",&a,&b,&c);
gets(s);
b1=solve(a,b,c,s);
gets(s);
scanf("%lf^%lf'%lf",&a,&b,&c);
gets(s);
a2=solve(a,b,c,s);
scanf("and %lf^%lf'%lf",&a,&b,&c);
gets(s);
b2=solve(a,b,c,s);
gets(s);
double xx=6875.0/2;
xx=xx*acos(sin(a1)*sin(a2)+cos(a1)*cos(a2)*cos(b1-b2));
printf("The distance to the iceberg: %.2lf miles.\n",xx);
if(dblcmp(xx-100+0.005)<0)
printf("DANGER!\n");
}
return 0;
}
return 0;
return k>0?1:-1;
}
int main()
{
double a,b,c;
double a1,b1,a2,b2;
while(gets(s))
{
gets(s);
gets(s);
scanf("%lf^%lf’%lf",&a,&b,&c);
gets(s);
a1=solve(a,b,c,s);
scanf(“and %lf^%lf’%lf”,&a,&b,&c);
gets(s);
b1=solve(a,b,c,s);
gets(s);
scanf("%lf^%lf'%lf",&a,&b,&c);
gets(s);
a2=solve(a,b,c,s);
scanf("and %lf^%lf'%lf",&a,&b,&c);
gets(s);
b2=solve(a,b,c,s);
gets(s);
double xx=6875.0/2;
xx=xx*acos(sin(a1)*sin(a2)+cos(a1)*cos(a2)*cos(b1-b2));
printf("The distance to the iceberg: %.2lf miles.\n",xx);
if(dblcmp(xx-100+0.005)<0)
printf("DANGER!\n");
}
return 0;
}