Title Translation
Sally Jones (Sally Jones) has more than a dozen travelers silver dollar. However, only eleven silver coin is true. A coin is counterfeit, even though its color and size of silver dollars with no real difference. The weight of counterfeit coins and other coins of different weight, but Sally does not know it is heavier than the real coins or light.
Fortunately, Sally had a friend borrowed her a very accurate balance sheet. Friends will allow Sally weighed three times to find counterfeit money. For example, if Sally and two coins each weighing scales balance, the two coins she knew it was right. Now, if Sally weighed
A real coin in the coin with respect to the third, unbalanced scales, third and Sally know coins are counterfeit, she may be placed on a balance which rise or fall to distinguish it in the light or heavy.
By carefully choosing her weighed, Sally can ensure that she will pinpoint three weighing coins.
Entry
The first line of the input integer n (n> 0), is used to specify the number of cases to follow. Three-row input in each case, each line were weighed once. Sally (Sally) identified by the letters A-L for each coin. Weighing two strings of information by letters and are given on one or even-numbered. The first letter string representing the balance of the coin on the left side; the second string, the right balance of the coin. (Sally always placed on the right side of the balance and the balance of the left side of the same number of coins.) The third position in the word will tell the right side of your balance is up, down or remain the same.
Ideas analysis
Ideas: one by one to determine whether each is true gold.
1, the day when the level is even, inside gold coins are true;
2, when the balance is not even, fake gold coins in the balance, and not there are true , while the balance in the gold mark. Divided into light and heavy.
3, when a gold coin is both light and heavy when, really.
The third point is also very important, such as the next set of data, even a very weak condition, we must up from two, down in to Tell on him with AB in the above time is the heavy side, and lighter below, that he is not a decisive factor.
ABCDEF GHIJKL up
ABHLEF GDIJKC down
CD HA even
The output is false coins.
Attach a set of data:
12
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even
AGHL BDEC even
JKI ADE up
J K even
ABCDEF GHIJKL up
ABC DEF even
I J down
ABCDEF GHIJKL up
ABHLEF GDIJKC down
CD HA even
A B up
B A down
A C even
A B up
B C even
DEFG HIJL even
ABC DEJ down
ACH IEF down
AHK IDJ down
ABCD EFGH even
AB IJ even
A L down
EFA BGH down
EFC GHD even
BA EF down
A B up
A C up
L K even
ACEGIK BDFHJL up
ACEGIL BDFHJK down
ACEGLK BDFHJI down
ACEGIK BDFHJL up
ACEGIL BDFHJK down
ACEGLK BDFHJI up
sample output
K is the counterfeit coin and it is light.
I is the counterfeit coin and it is heavy.
I is the counterfeit coin and it is light.
L is the counterfeit coin and it is light.
B is the counterfeit coin and it is light.
A is the counterfeit coin and it is heavy.
A is the counterfeit coin and it is light.
L is the counterfeit coin and it is heavy.
A is the counterfeit coin and it is light.
A is the counterfeit coin and it is heavy.
L is the counterfeit coin and it is light.
K is the counterfeit coin and it is heavy.
#include<iostream>
#include<map>
#include<string>
#include<string.h>
using namespace std;
int coins[12];//条件码 0：待确定 1：真的 2：未曾判断 所有硬币最后只能有一个待确定，就是假的
int cnt = 0, weight[12];//已经确认为真的数目，weight：如果是假的，是重:1还是轻:-1。
int main() {
int n; cin >> n;
string s1, s2, s3;
while (n--) {
for (int i = 0; i < 12; i++)coins[i] = 2;
memset(weight, 0, sizeof(weight));
for (int i = 0; i < 3; i++) {
cin >> s1 >> s2 >> s3;
if (s3[0] == 'e') {
for (int j = 0; j < s1.size(); j++)coins[s1[j] - 'A'] = 1;
for (int j = 0; j < s2.size(); j++)coins[s2[j] - 'A'] = 1;
}
else if (s3[0] == 'u') {//右边轻
for (int j = 0; j < s1.size(); j++)
{//不是真的就可能是假的
if (coins[s1[j] - 'A'] == 2)coins[s1[j] - 'A'] = 0, weight[s1[j] - 'A'] = 1;
else if (weight[s1[j] - 'A'] != 1) {//之前称过，而且不是重的一方
coins[s1[j] - 'A'] = 1;
}
}
for (int j = 0; j < s2.size(); j++) {
if (coins[s2[j] - 'A'] == 2)coins[s2[j] - 'A'] = 0, weight[s2[j] - 'A'] = -1;
else if (weight[s2[j] - 'A'] != -1) {
coins[s2[j] - 'A'] = 1;
}
}
for (int j = 0; j < 12; j++) {
int a = s1.find(j + 'A'), b = s2.find(j + 'A');
if (a == -1 && b == -1)
coins[j] = 1;//不在假的里面就是真的
}
}
else {//左边轻
for (int j = 0; j < s1.size(); j++)
{//不是真的就可能是假的
if (coins[s1[j] - 'A'] == 2)coins[s1[j] - 'A'] = 0, weight[s1[j] - 'A'] = -1;
else if (weight[s1[j] - 'A'] != -1) {
coins[s1[j] - 'A'] = 1;
}
}
for (int j = 0; j < s2.size(); j++) {
if (coins[s2[j] - 'A'] == 2)coins[s2[j] - 'A'] = 0, weight[s2[j] - 'A'] = 1;
else if (weight[s2[j] - 'A'] != 1) coins[s2[j] - 'A'] = 1;
}
for (int j = 0; j < 12; j++) {
int a = s1.find(j + 'A'), b = s2.find(j + 'A');
if (a == -1 && b == -1)
coins[j] = 1;//不在假的里面就是真的
}
}
}
for (int i = 0; i < 12; i++) {
if (coins[i] == 0) {
char s = i + 'A';
if (weight[i] == 1) s1.assign("heavy");
else s1.assign("light");
cout << s << " is the counterfeit coin and it is " << s1 << "." << endl; break;
}
}
}
}
