uestc 969 易位法字符串解密

字符串解密"/>

uestc 969 易位法字符串解密

题意：chinese

思路：就是按照题意模拟

题目链接：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cctype>
using namespace std;const int maxn = 1005;char a[30], s[1005], temp[1005];
int ran[30];
vector<node> v;
vector<char> v1;
struct node
{char s[1005];int val;bool operator < (const node &rhs) const{return val < rhs.val;}
};int main()
{int t;scanf("%d", &t);while (t--){v.clear(), v1.clear();scanf("%s%s", a, s);int len1 = strlen(a), len2 = strlen(s);int res = len2 / len1;for (int i = 0; i < len1; i++){ran[a[i] - 'A'] = i;v1.push_back(a[i]);}sort(v1.begin(), v1.end());node e;for (int i = 0; i < len1; i++){strncpy(e.s, s + i * res, res);e.val = ran[v1[i] - 'A'];v.push_back(e);}sort(v.begin(), v.end());int cnt = 0;while (cnt < res){for (int i = 0; i < (int)v.size(); i++)printf("%c", toupper(v[i].s[cnt]));cnt++;}printf("\n");}return 0;
}