HDU 4492 Mystery （模拟）

Mystery

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 548    Accepted Submission(s): 275

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

41MAC31 1 12IW2C0NP3OS 1RLDFA220  3  3  -3  7  -8  2  7  -4  3  8  7  4  1  1  -4  5  2  5  -6  -3  -43G.IETSNPRBU172  4  5  -6  -1  -3  -2  -4  -4  1  -1  5  -3  4  1  -2  44PIBN MRDSYEO16-4  4  -1  4  5  3  -5  4  -3  -3  -2  -5  -5  -3  1  3

 

Sample Output

1 ACM2 ICPC 2013 WORLD FINALS3 IN ST. PETERSBURG4 SPONSORED BY IBM

 

Source

Greater New York 2012

 

Recommend

liuyiding


题意：

给你一字符串和一些整数，这些整数表示相对于当前位置的位移，并输出该位所对应的字符。
起始位置位于0。样例3是有一点问题的，因为输入中并没有空格。

思路：

1.直接按题意模拟
2.读入字符串时，建议使用fgets，不易发生内存越界，堆栈溢出等问题，较安全，因为它规定了数组长度。
3.输入时回车符最好别使用'\n',一开始就错这，可以使用getchar（），或者scanf（“%d%*c”，&num）；

/*************************************************************************> File Name: 4492.cpp> Author: BSlin> Mail:  > Created Time: 2013年10月06日 星期日 18时00分41秒 ************************************************************************/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <algorithm>#include <iterator>#include <vector>#include <map>#include <set>#include <stack>#include <queue>#define MP make_pair#define INF (1<<30)#define PI acos(-1.0)#define esp 1e-8const int dx[4]={0,0,0,0};using namespace std;#define read freopen("in.txt","r",stdin)#define write freopen("out.txt","w",stdout)#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)#define LL __int64#define LLS "%" "I" "6" "4" "d"#else#define LL long long#define LLS "%" "l" "l" "d"#endifint main(int argc, char** argv) {    //read;    int t,num,n,pos,len;    char str[100];    int a[520];    scanf("%d",&t);    while(t--) {        scanf("%d",&num);          //用getchar()，或%*c，别用\n        getchar();        fgets(str,100,stdin);      //fgets 挺好        len = strlen(str);        len --;        scanf("%d",&n);        for(int i=0; i<n; i++) {            scanf("%d",&a[i]);        }        pos = 0;        printf("%d ",num);        for(int i=0; i<n; i++) {            pos = (pos + a[i] + len) % len;            printf("%c",str[pos]);        }        printf("\n");    }    return 0;}