对于输入字符串数组的处理，三维数组

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

1
8 5 2
94 62 80 00 24
79 11 07 80 03
32 66 12 89 48
57 82 36 69 32
54 66 91 54 93
12 14 37 92 83
52 70 33 89 95
59 91 01 80 69
A4B7D1
E5C3B2

Sample Output

Case 1:
577000
931211

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

对于字符串数组的处理，可以使用三维数组

具体的见代码

#include 
#include 
#include 

using namespace std;

char map[15][15][15];
char str[105];
int main () {
	int cas;
	memset(map, 0, sizeof(map));
	scanf("%d", &cas);
	for (int t = 1; t <= cas; t ++) {
		int n, m, q;
		scanf("%d%d%d", &n, &m, &q);
		for (int i = 1; i <= n; i ++)
			for (int j = 0; j < m; j ++) {
				scanf("%s", &map[i][j]);
			}
			printf("Case %d:\n", t);
			for (int i = 0; i < q; i ++) {
				scanf("%s", str);
				int len = strlen(str);
				for(int i = 0; i < len; i += 2) {
					printf("%s", map[str[i + 1] - '0'][str[i]-'A']);
				}
				printf("\n");
			}
	}
	return 0;
}