FZU 2034-Password table(模拟)

Problem 2034 Password table

Accept: 424    Submit: 816

Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Do you know password table? A password table is used to protect the security of the online account. When a user needs to login to his/her online account or pay for money, he/she may need to fill in the password according to his/her own password table. For
example, if one user has the following table and the online system requires him/her to input the password of “A4B7D1”, he/she may input “577000”.

 Input

The first line of the input contains an integer T(T≤100), indicating the number of test cases. The first line of each case contains three numbers n(5≤n≤9), m(5≤m≤9) and q(1≤q≤100) representing the length and width of the password table, and the total number
of queries (Look at the password table above, its length is 8 and its width is 5). Each of the following n lines contains m numbers, representing the given password table.

Then, each of the following q lines represents a query with the format of "L1D1L2D2L3D3" (L1, L2 and L3 are letters. D1, D2 and D3 are digits), just like the example in paragraph one.

 Output

For each test case, print a line containing the test case number (beginning with 1). For each query in one test case, please output its corresponding password in one line.

 Sample Input

18 5 294 62 80 00 2479 11 07 80 0332 66 12 89 4857 82 36 69 3254 66 91 54 9312 14 37 92 8352 70 33 89 9559 91 01 80 69A4B7D1E5C3B2

 Sample Output

Case 1:577000931211

 Source

2011年全国大学生程序设计邀请赛（福州）

题意：题目的意思，给出n*m的数组，然后给出e行cas、行数从1编号到n，列数从A到。。给你字符串，求出对应的表中的数据。

用三维数组水水的模拟一下就好了

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double pi= acos(-1.0);
char a[10][10][1010];
int main()
{
    int T,n,m,q,i,j;
    int icase;
    char str[110];
    scanf("%d",&T);
    for(icase=1;icase<=T;icase++){
        scanf("%d %d %d",&n,&m,&q);
        for(i=1;i<=n;i++)
            for(j=0;j<m;j++)
            scanf("%s",&a[i][j]);
        printf("Case %d:\n", icase);
        while(q--){
            scanf("%s",str);
            int len=strlen(str);
            for(i=0;i<len;i+=2){
                printf("%s",a[str[i+1]-'0'][str[i]-'A']);
            }
            puts("");
        }
        }
    return 0;
}