其实一贯都是用string来着。。

其实后来看刘汝佳的书又喜欢上了fgets然后一路过关斩将来着。。


这次觉得题水就用了gets啊啊啊啊

然后就WA了9次啊有木有！！！


H. Decode

<center style="font&#45;family&#58;monospace&#59;font&#45;size&#58;14px&#59;"> TimeLimit: 1s   MemoryLimit: 64M </center>

Description

Bruce Force has had an interesting idea how to encode strings. The following is the description of how the encoding is done:
Let x 1 ,x 2 ,...,x n  be the sequence of characters of the string to be encoded.
1.Choose an integer  m and  n pairwise distinct numbers p 1,p 2,...,p n from the set { 1,  2, ...,  n} (a permutation of the numbers  1 to  n).
2.Repeat the following step  m times.
3.For  1 ≤ i ≤  n set y i to x pi, and then for  1 ≤ i ≤  n replace x i by y i.
For example, when we want to encode the string "hello", and we choose the value  m = 3  and the permutation  2, 3, 1, 5, 4 , the data would be encoded in 3 steps: "hello" - > "elhol" -> "lhelo" -> "helol".
Bruce gives you the encoded strings, and the numbers  m  and p 1 , ..., p n  used to encode these strings. He claims that because he used huge numbers  m  for encoding, you will need a lot of time to decode the strings. Can you disprove this claim by quickly decoding the strings?

Input

The input contains several test cases. Each test case starts with a line containing two numbers nand m (1 ≤ n ≤ 80, 1 ≤ m ≤ 109). The following line consists of n pairwise different numbers p1,...,pn (1 ≤ pi ≤ n). The third line of each test case consists of exactly n characters, and represent the encoded string. The last test case is followed by a line containing two zeros.

Output

For each test case, print one line with the decoded string.

SampleInput

5 3
2 3 1 5 4
helol
16 804289384
13 10 2 7 8 1 16 12 15 6 5 14 3 4 11 9
scssoet tcaede n
8 12
5 3 4 2 1 8 6 7
encoded?
0 0

SampleOutput

hello
second test case
encoded?


不说啥了，就是mod一个周期就行。getline的代码在最下面，纪念一下死的多坑。


#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <ctime>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#define LL long long

using namespace std;

const int INF = 999999999;
const double eps = 1e-6;

int T,I,n,m,t;
LL tt, sum;
int a[1111];
int f[1111];
int dp[888][888];
char s[888];
char ans[888];
char ss[888];

int so(int v){
    if (a[v] == v ) return 1;
    int t = a[v];
    int res = 1;
    while (t!=v){
        t = a[t];
        res++;
    }
    return res;
}

int too(int v,int k){
    for (int i(1);i<=k;i++){
        v = a[v];
    }
    return v;
}

int main(){
    while ( fgets(ss, INF, stdin) ) {
        sscanf(ss, "%d%d",&n,&m);
        if (n==0 && m==0) break;

        fgets(ss, INF, stdin);
        char *p = strtok(ss, " ");
        for (int i(1);i<=n;i++) {
            sscanf(p, "%d", &a[i]);
            p = strtok(NULL, " ");
            f[i] = 0;
        }

        for (int i(1);i<=n;i++) {
            f[i] = so(i);
        }

        fgets(s+1, INF, stdin);

        for ( int i(1); i<=n; i++) {
            ans[too(i, m%f[i] )] = s[i];
        }
        ans[n+1] = 0;
        printf("%s\n",ans+1);
    }
    return 0;
}



#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
using namespace std;
const int MAXN = 888;
int n, m;
int a[MAXN], f[MAXN];
char ans[MAXN], s[MAXN], ch;

int to(int v, int w){
    while( w-- )    v = a[v];
    return v;
}

int main(){    
    while( EOF != scanf("%d %d\n", &n, &m) ){
        if( !n && !m )  break;
        //getchar();
        for(int i=1; i<=n; i++)
            cin>>a[i];//scanf("%d", &a[i]), getchar();
        cin.getline(s+1, MAXN);
        cin.getline(s+1, MAXN);
        for(int i=1; i<=n; i++)
            if( a[i]==i )   f[i] = 1;
            else{
                int t=a[i], cnt=1;
                while( t!=i )    t=a[t],     cnt++;
                f[i] = cnt;
            }
        for(int i=1; i<=n; i++)
            ans[ to(i, m%f[i]) ] = s[i];
        ans[n+1] = 0;
        cout<<ans+1<<endl;
    }
    return 0;
}

真切感慨以后在时间充裕的时候能用C++就用C++。别自作多情的装逼用c啥啥的