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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
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