8.2 Jizhong training Day2

T1 cuisine

Description

　　Food is very tasty dish of meaning, food is the most important to choose a good raw material. 
　　There are N kinds of ingredients, each of both acidity and S B Bitterness two attributes, when selecting a variety of materials, a total acidity of the acidity of the product of each ingredient, the total degree of bitterness Bitterness of each ingredient with. 
　　As you know, neither food nor bitter acid, as a raw material to ensure that the selected total acidity and total pain of the minimum absolute difference. 
　　Because food is not only water, so you must select at least one meal.

Input

　　Input of the first row contains an integer N (1 <= N <= 10), indicates the number of kinds of raw materials. 
　　Next N lines contains two integers separated by a space, respectively, of the acidity and bitter. 
　　If the input data to ensure that all the raw materials are selected, total acidity and total pain of no more than 10 ^ 9.

Output

　　Minimum difference between the total acidity and total output of bitter.

Examination room ideas / positive solution

       what? 1 <= N <= 10? Dfs! How violence, how ever (no brain)!

Code

#include<stdio.h>
#include<algorithm>
using namespace std;

int n,ans;
int book[15];
struct thm
{
    int a;
    int b;
}sz[15];

void Work(int nowa,int nowb)
{
    ans=min(abs(nowa-nowb),ans);
    for(int i=1;i<=n;i++)
    {
        if(book[i])
            continue;
        book[i]=1;
        Work(nowa*sz[i].a,nowb+sz[i].b);
        book[i]=0;
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&sz[i].a,&sz[i].b);
    ans=2e9;
    for(int i=1;i<=n;i++)
    {
        book[i]=1;
        Work(sz[i].a,sz[i].b);
        book[i]=0;
    }
    printf("%d",ans);
    return 0;
}

T2 fetch game

Description

　　Alice wants Bob to accompany him to see "the Great Tangshan Earthquake", but because Bob is a very emotional person, tears do not want to be afraid, but I am sorry this as a reason to refuse, he proposed to play a game.
　　N positive integers circle, rules are as follows:
　　• two players take turns fetching;
　　• the beginning just get a player can take any number x;
　　• the beginning of the current player can only take x from the second step (the player just take a number) left and right sides of adjacent numbers;
　　• take complete until all the numbers, the game is over;
　　• get more odd number of wins.
　　Bob in order to show generosity, let Alice first take, but he forgot himself and Alice are very smart people, Alice now ask you to help him calculate how many emulated the first step in making the final victory.

Input

　　The first row contains an integer N (1 <= N <= 100), it represents the number of values. The second line contains N positive integers, each number between 1 and 1000, any two numbers different from each other.

Output

　　The first step in output Alice how many emulated.

Sample Input

Input 1: 
3 
3 1. 5 

Input 2: 
. 4 
1 2 3. 4 

input 3: 
. 8 
. 4. 9 10. 8. 5 2 1. 7

Sample Output

Output 1: 
3 

Output 2: 
2 

Output 3: 
5

Examination room ideas / positive solution

emmmm This is obviously a question of range DP (I think that is the most difficult question 4.), I spent a half-hour exam to push this problem, and finally launch, but there is some small mistakes in the handling of the border, only 50 (tears ran).

Provided f1 [i] [j] denotes the current length i, the odd and the position obtained when the first element of j and the upper hand is optimal.

    f2 [i] [j] represents the length of the current i, and the odd position of the first element when j is obtained when the optimal flip.

So we is not difficult to draw: If f2 [i-1] [j + 1] + sz [j]% 2> f2 [i-1] [j] + sz [i + j-1]% 2, then f1 [i] [j] = f2 [i-1] [j + 1] + sz [j]% 2, f2 = f1 [i-1] [j + 1], otherwise, f1 [i] [j] = f2 [i-1] [j] + sz [i + j-1]% 2, f2 [i] [j] = f1 [i-1] [j]. (Now the upper hand, will be carried round the access number for the remaining flip-1 i, but now the upper hand FLAC Similarly i-1 for the remaining number.)

so, the core code is:

if((f2[i-1][j+1]+sz[j]%2>f2[i-1][j]+sz[i+j-1]%2) || (f2[i-1][j+1]+sz[j]%2==f2[i-1][j]+sz[i+j-1]%2 && f1[i-1][j+1]<f1[i-1][j]))
	f1[i][j]=f2[i-1][j+1]+sz[j]%2,f2[i][j]=f1[i-1][j+1];
else
	f1[i][j]=f2[i-1][j]+sz[i+j-1]%2,f2[i][j]=f1[i-1][j];

　　(Remember the process boundary)

Code

#include <stdio.h> 
#include <algorithm>
 the using  namespace STD; 

int n-, L, R & lt, ANS;
 int SZ [ 110 ];
 int F1 [ 110 ] [ 110 ]; / * F1 [I] [J]: when only the i-th element, and the first element position j, just get optimal value * / 
int F2 [ 110 ] [ 110 ]; / * F2 [i] [J]: when only the i-th element, and the first element position to an optimal value j, FLAC * / 

int main () 
{ 
    int a, B; 
    Scanf ( " % D " , & n-);
     for ( int I = . 1;i<=n;i++)
        scanf("%d",&sz[i]),sz[i+n]=sz[i];
    for(int i=1;i<=2*n;i++)
        f1[1][i]=sz[i]%2;
    for(int i=2;i<n;i++)
        for(int j=1;j<=n;j++)
        {
            a=j+1>n ? (j+1)%n : j+1;
            b=i+j-1>n ? (i+j-1)%n : i+j-1;
            if((f2[i-1][a]+sz[j]%2>f2[i-1][j]+sz[b]%2) || (f2[i-1][a]+sz[j]%2==f2[i-1][j]+sz[b]%2 && f1[i-1][a]<f1[i-1][j]))
                f1[i][j]=f2[i-1][a]+sz[j]%2,f2[i][j]=f1[i-1][a];
            else
                f1[i][j]=f2[i-1][j]+sz[b]%2,f2[i][j]=f1[i-1] [J]; 
        } 
    for ( int I = . 1 ; I <= n-; I ++ ) 
    { 
        A = I + . 1 > n-(I +? . 1 )% n-: I + . 1 ;   // bloggers is here forget sentence boundaries, cautionary 
        IF (F2 [N- . 1 ] [A] + SZ [I]% 2 > F1 [N- . 1 ] [A]) 
            ANS ++ ; 
    } 
        
    the printf ( " % D " , ANS);
     return  0 ; 
}

T3 delete

Description

　　Alice on chemical class and distracted, he first drew a line 3 of Form N columns, then the first row of numbers 1 to N fill in the form, to ensure that each number appears only once, while he also filled two rows numbers 1 to N, but not limit the number of occurrence of each number.
　　Alice now want to delete several columns so that each row after row of identical sequence finished, the program requires a minimum of computing how many columns to delete.

Input

　　The first row contains an integer N (1 <= N <= 100000), the number of columns of the table.
　　The next three lines each line contains N integers, each a number between 1 to N, the number of the first row and different from each other.

Output

　　Output requires a minimum number of columns deleted.

Sample Input

Input 1: 
. 7 
. 5. 4. 3 2 1. 6. 7 
. 5. 5 1 1. 3. 4. 7 
. 3. 7 1. 4. 5. 6 2 

Input 2: 
. 9 
1. 3. 5. 9. 8. 6 2. 4. 7 
2 1. 5. 6. 4. 9. 3. 4. 7 
. 3. 5 1 986287

Sample Output

Output 1: 
4 

Output 2: 
2

Hint

[Explain] Sample
　　Example 1, Alice needs to delete four of 2,4,6,7, and each row is drained sequence 1,3,5.
[Data] range
　　of 40% of the data N <= 100
　　70% of the data N <= 10000

 

Examination room ideas / positive solution

Water problem, simulated violence

 

Code

#include<cstdio>
#include<algorithm>
using namespace std;
int n,kg,ans,book[100010];
int a[100010],b[100010],c[100010];
int cb[100010],cc[100010];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]),cb[b[i]]++;
    for(int i=1;i<=n;i++)
        scanf("%d",&c[i]),cc[c[i]]++;
    while(kg==0)
    {
        kg=1;
        for(int i=1;i<=n;i++)
        {
            if(book[i]==0 && (cb[a[i]]<=0 || cc[a[i]]<=0))
            {
                kg=0;
                ans++;
                book[i]=1;
                cb[b[i]]--;
                cc[c[i]]--;
            }
        }
    }
    printf("%d",ans);
    return 0;
}

T4 interval

Description

　　Alice received some very special birthday gift: interval. Even if it is boring, Alice was able to come up with a lot of games on intervals, one of which is, Alice longest sequence selected from different sections, where each section must meet the gift, the other in the sequence must contain at each interval an interval.
　　Program calculated maximum length sequence.

Input

　　The first line of the input file contains an integer N (1 <= N <= 100000), indicates the number of sections.
　　Next N lines of two integers A and B depict a section (1 <= A <= B <= 1000000).

Output

　　The maximum length sequence output condition satisfied.

Sample Input

Input 1: 
3 
3. 4 
2. 5 
1. 6 

Input 2: 
. 5 
10 30 
20 is 40 
30 50 
10 60 
30 40 

Input 3: 
. 6 
1. 4 
1. 5 
1. 6 
1. 7 
2. 5 
3. 5

Sample Output

Output 1: 
4 

Output 2: 
2

Hint

[Explain] Sample
　　Example 3 can be found in sequence interval length is 5: [1,7], [1,6], [1,5], [2,5], [3,5]

Examination room ideas

No time to play, emmmmm .....

Clarification: there is no clear title, obviously it is to find a few covered range. So obvious as long as the left point range to do it again sort, then it becomes the subject does not rise up to demand the sequence.

Code

#include<stdio.h>
#include<algorithm>
using namespace std;

int n,Len,a;
int s[100010];
struct thm
{
    int l;
    int r;
}sz[100010];

bool cmp(thm a,thm b)
{
    if(a.l==b.l)
        return a.r>b.r;
    return a.l<b.l;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&sz[i].l,&sz[i].r);
    sort(sz+1,sz+1+n,cmp);
    Len=1;
    for(int i=1;i<=n;i++)
    {
        if(s[Len]<=-sz[i].r)
            s[++Len]=-sz[i].r;
        else
        {
            a=upper_bound(s+1,s+1+n,-sz[i].r)-s;
            s[a]= -sz[i].r;
        }
    }
    printf("%d",Len);
    return 0;
}

to sum up

Today partial simple topic, but did not test well, the interval DP equation push for too long, it seems not enough skilled, much brush title.