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UVA 306 Cipher

题意 :lucky cat里有翻译。英文也比较好懂。

很容易发现有周期然后就拍就好了

注意每组数据后边都有空行 包括最后一组。一开始以为最后一组没有空行。唉。。

#include <map>#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <stack>#include <queue>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <climits>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define PI 3.1415926535897932626using namespace std;int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}int t[210];//周期int N,pos[210],tmp[210];int k,T,res;char input[210];int ans[210];void dfs(int cur,int depth){    if (cur == res) {T = depth; return ;}    dfs(pos[cur],depth + 1);}void calcu(){    for (int i = 1; i <= N; i++)    {        res = i;        dfs(pos[i],1);        t[i] = T ;    }}int main(){    //freopen("sample.txt","r",stdin);    bool first = false;    while (scanf("%d",&N)!=EOF)    {        if (N == 0) break;        for (int i = 1; i <= N; i++) scanf("%d",&pos[i]);        calcu();        //for (int i = 1; i <= N; i++) printf("%d ",t[i]);putchar(‘\n‘);        while (scanf("%d",&k) != EOF)        {          if (k == 0) break;          getchar();          gets(input+1);          int len = strlen(input + 1);          for (int i = len + 1; i <= N; i++)             input[i] =  ;          input[N + 1] = \0;          //for (int i = 1; i <= N; i++) printf("%d ",tmp[i]);puts("");          char output[210];          for (int i = 1; i <= N; i++) tmp[i] = i;          for (int i = 1; i <= N; i++)          {              int tp = k % t[i];              for (int j = 0; j < tp; j++)                    tmp[i]=pos[tmp[i]];          }          for (int i = 1; i <= N;i++) output[tmp[i]] = input[i];          for (int i = 1; i <= N;i++)            printf("%c",output[i]);          putchar(\n);        }        putchar(\n);    }    return 0;}

 

UVA 306 Cipher