Mystery

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 141    Accepted Submission(s): 75

Input The first line of input contains a single integer P, (1 <= P <= 1000), which is the number of data sets that follow. Each data set consists of several lines. Each data set should be processed identically and independently.

The first line of each data set contains an integer D which is the data set number. The second line contains no more than the 93 distinct printable ASCII characters. The third line contains an integer, N (1 <= N <=512 ), which is the number of integers on the next (fourth) line of the dataset. Each integer on the fourth line is in the range -X to X where X is the number of characters on the second line minus 1.   Output For each data set there is one correct line of output. It contains the data set number (D) followed by a single space, followed by a string of length N made of the characters on the second line of the input data set.   Sample Input 4 1 MAC 3 1 1 1 2 IW2C0NP3OS 1RLDFA 22 0 3 3 -3 7 -8 2 7 -4 3 8 7 4 1 1 -4 5 2 5 -6 -3 -4 3 G.IETSNPRBU 17 2 4 5 -6 -1 -3 -2 -4 -4 1 -1 5 -3 4 1 -2 4 4 PIBN MRDSYEO 16 -4 4 -1 4 5 3 -5 4 -3 -3 -2 -5 -5 -3 1 3   Sample Output 1 ACM 2 ICPC 2013 WORLD FINALS 3 IN ST. PETERSBURG 4 SPONSORED BY IBM   Source
Greater New York 2012   Recommend liuyiding    
刷了下HDU的最后一题，竟然是非常水的水题，MARK下

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
char str[100];
int main()
{
    int T;
    int D;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%*c",&D);//跳过后面的一个换行符
        printf("%d ",D);
        cin.getline(str,100);
        int len=strlen(str);
        int n;
        int t=0;
        scanf("%d",&n);
        int x;
        while(n--)
        {
            scanf("%d",&x);
            t+=x;
            t=(t+len)%len;
            printf("%c",str[t]);
        }
        printf("\n");
    }
    return 0;
}