POJ 1027 Cipher(置换)

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题意：第i个移动到a[i]。问k次之后的字符串。

思路：置换求循环节。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 2e2 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, k;char s[N];bool vis[N];struct node {int num, len;} a[N];int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    while(si(n) != EOF) {    for(int i = 1; i <= n; i ++) si(a[i].num);    mem(vis, false);    for(int i = 1; i <= n; i ++) {    int cnt = 0, idx = i;    while(!vis[idx]) {    vis[idx] = 1;    cnt ++;    idx = a[idx].num;    }    idx = i;    for(int j = 0; j < cnt; j ++) {    a[idx].len = cnt;    idx = a[idx].num;    }    }    while(si(k), k) {    gets(s);    char ans[N];    int sz = strlen(s);    for(int i = sz; i <= n; i ++) s[i] = ' '; s[n+1] = '\0';    for(int i = 1; i <= n; i ++) {    int cnt = k % a[i].len, idx = i;    for(int i = 0; i < cnt; i ++) idx = a[idx].num;    ans[idx] = s[i];    } ans[n+1] = '\0';    puts(ans + 1);    }    puts("");}        return 0;}

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