ACM_Mystery

Mystery

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

No Description

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
解题思路：简单模拟一下n次访问字符串中的字符，如果访问的下标（每次将x叠加）小于0，要加上第二行字符串的长度len，再取余len输出当前被访问的字符即可（第三组测试样例应该是错的），水过！
AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int p,d,n,x,j,k,len;char ch[95],str[550];cin>>p;
 5     while(p--){
 6         cin>>d;getchar();gets(ch);
 7         j=k=0;len=strlen(ch);memset(str,'',sizeof(str));
 8         cin>>n;
 9         while(n--){
10             cin>>x;k+=x;
11             if(k<0)k+=len;
12             str[j++]=ch[k%len];
13         }
14         cout<<d<<' '<<str<<endl;
15     }
16     return 0;
17 }