Mystery

Mystery
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB
Total submit users: 19, Accepted users: 18
Problem 12606 : No special judgement
Problem 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 


Sample Input
7 
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 
5
D^obV@k"W*B&#]4!NcF$'lj%(d6XG5fi<Hxz7)2Lt~=8aQuvh}r_m+C9eI`-.>EwYyngZRsJKpqO{[\U|MPS,;T?031/A 
93
-1 11 44 39 -31 -44 10 5 24 14 1 -33 42 28 -34 7 -37 24 14 3 -7 18 4 19 37 4 20 2 41 -42 18 15 -3 10
7 12 -11 -41 14 8 31 -26 37 -19 -17 -9 -16 15 31 14 29 -22 1 -24 20 -30 6 1 16 -29 31 -30 6 17 -43 -
10 7 7 4 -22 10 -2 15 13 14 2 6 -17 34 -27 28 29 -28 2 33 -13 -15 6 -31 24 41 29 26
6
I
1
0
7
I
13
0 0 0 0 0 0 0 0 0 0 0 0 0
Sample Output
1 ACM
2 ICPC 2013 WORLD FINALS
3 INST. PETERSBURG
4 SPONSORED BY IBM
5 ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz~'!@#$%^&*()_+-={}[]\|'";/.,<>?
6 I
7 IIIIIIIIIIIII
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 using namespace std;
 5 char a[200];
 6 int b[800];
 7 int main(void){
 8 #ifndef ONLINE_JUDGE
 9   freopen("in", "r", stdin);
10 #endif
11   int t; scanf("%d", &t);
12   while (t--){
13     int s; scanf("%d", &s);
14     getchar();
15     gets(a);
16     int n; scanf("%d", &n);
17     for (int i = 0; i < n; ++i) scanf("%d", b+i);
18     printf("%d ", s);
19     int po = 0, len = strlen(a);
20     for (int i = 0; i < n; ++i) {
21       po = po + b[i];
22       if (po >= len) po = po % len;
23       else if (po < 0) po = (po + len);
24       printf("%c", a[po]);
25     }
26     printf("\n");
27   }
28 
29   return 0;
30 }

题目不难,就是卡了很久,最后才发现,原来是输入的文件里面写错了,多打了一个空格……次奥……

原文地址:https://www.cnblogs.com/liuxueyang/p/2952783.html