題意;給你兩個6行5列的密碼矩陣,找出滿足下列條件的“密碼”:密碼中的每一個字母都在兩個矩陣的對應列中出現,求字典序第k大的的密碼,如果沒有輸出NO。
題意:直接暴力瞎搞一下就好了。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define MAXN 10
#define MAXE 400
#define INF 100000000
#define MOD 10001
#define LL long long
#define pi 3.14159
using namespace std;
string matrix1[MAXN];
string matrix2[MAXN];
set<char> Set[MAXN];
int main() {
std::ios::sync_with_stdio(false);
int T;
cin >> T;
for (int kase = 1; kase <= T; ++kase) {
int k;
for (int i = 0; i < MAXN; ++i) {
Set[i].clear();
}
cin >> k;
for (int i = 0; i < 6; ++i) {
cin >> matrix1[i];
}
for (int i = 0; i < 6; ++i) {
cin >> matrix2[i];
}
set<char> s;
for (int i = 0; i < 5; ++i) {
s.clear();
for (int j = 0; j < 6; ++j) {
s.insert(matrix1[j][i]);
}
for (int j = 0; j < 6; ++j) {
if (s.count(matrix2[j][i])) {
Set[i].insert(matrix2[j][i]);
}
}
}
if (Set[0].size() * Set[1].size() * Set[2].size() * Set[3].size() * Set[4].size() < k) {
cout << "NO\n";
} else {
int num1 = (int) Set[1].size();
int num2 = (int) Set[2].size();
int num3 = (int) Set[3].size();
int num4 = (int) Set[4].size();
vector<int> vec;
int pos = k / (num1 * num2 * num3 * num4);
k %= (num1 * num2 * num3 * num4);
if (k == 0) {
vec.push_back(pos - 1);
k = num1 * num2 * num3 * num4;
} else {
vec.push_back(pos);
}
pos = k / (num2 * num3 * num4);
k %= (num2 * num3 * num4);
if (k == 0) {
vec.push_back(pos - 1);
k = num2 * num3 * num4;
} else {
vec.push_back(pos);
}
pos = k / (num3 * num4);
k %= (num3 * num4);
if (k == 0) {
vec.push_back(pos - 1);
k = num3 * num4;
} else {
vec.push_back(pos);
}
pos = k / num4;
k %= num4;
if (k == 0) {
vec.push_back(pos - 1);
k = num4;
} else {
vec.push_back(pos);
}
pos = k;
vec.push_back(pos - 1);
string str = "\0";
for (int i = 0; i < 5; ++i) {
int ans = 0;
for (set<char>::iterator it = Set[i].begin(); it != Set[i].end(); ++it) {
if (ans == vec[i]) {
str += *it;
break;
}
ans++;
}
}
cout << str << endl;
}
}
return 0;
}
/*
3
1
AYGSU
DOMRA
CPFAS
XBODG
WDYPK
PRXWO
CBOPT
DOSBG
GTRAR
APMMS
WSXNU
EFGHI
5
AYGSU
DOMRA
CPFAS
XBODG
WDYPK
PRXWO
CBOPT
DOSBG
GTRAR
APMMS
WSXNU
EFGHI
64
FGHIJ
EFGHI
DEFGH
CDEFG
BCDEF
ABCDE
WBXDY
UWYXZ
XXZFG
YYFYH
EZWZI
ZGHIJ
*/