Password UVA - 1262

按照紫书的思路，首先找出对应的每列相同的字母，将这些字母分别存储之后，计算对应的数量，然后输出即可，具体实现见如下代码：

#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<deque>
#include<functional>
using namespace std;

int main(){
int T,K;
cin >> T;
while (T--){
cin >> K;
char first[6][5];
char second[6][5];
for (int i = 0; i < 6; i++){
for (int j = 0; j < 5; j++){
cin >> first[i][j];
}
}
for (int i = 0; i < 6; i++){
for (int j = 0; j < 5; j++){
cin >> second[i][j];
}
}
set<char> v[5];
for (int j = 0; j < 5; j++){
for (int ind1 = 0; ind1 < 6; ind1++){
for (int ind2 = 0; ind2 < 6; ind2++){
if (first[ind1][j] == second[ind2][j]){
v[j].insert(first[ind1][j]);
break;
}
}
}
}
int amount[6];
bool flag = false;
amount[5] = 1;
for (int i = 4; i >= 0; i--){
amount[i] = amount[i + 1] * v[i].size();
if (amount[i] == 0){
flag = true;
break;
}
}
if (K> amount[0]||flag){
cout << "NO\n";
continue;
}
int cur = 0;
set<char>::iterator j;
for (int i = 0; i < 5; i++){
for (j = v[i].begin(); j != v[i].end(); j++){
if (cur + amount[i + 1] < K){
cur += amount[i + 1];
}
else break;
}
cout << *j;
}
cout << endl;
}
return 0;
}

/*
3
ABDFK
OCOOO
ODOOO
OOOOO
OOOOO
OOOOO
ABDFK
ECEEE
EDEEE
EEEEE
EEEEE
EEEEE
*/