Uva 1368-DNA Consensus String

本题只有四个字符，用常量数组存储它们
#include <cstdio>
#include <cstring>

char a[100][1500];
int b[10];

int main()
{
    int cnt, m, n, hamming= 0,maxum = 0;
    char c[10] ="ACGT";
    scanf("%d",&cnt);
    while(cnt--)
    {
        scanf("%d%d",&m,&n);
        getchar();
        for(int i = 0; i < m; i++)
          {
              for(int j = 0; j < n;j++)
                 scanf("%c",&a[i][j]);
                    getchar();
          }
        for(int j = 0; j < n; j++)
         {
             memset(b,0,sizeof(b));
             for(int i = 0; i < m; i++)
            {
             if(a[i][j] == ‘A‘)
                b[0]++;
             if(a[i][j] == ‘C‘)
                b[1]++;
             if(a[i][j] == ‘G‘)
                b[2]++;
             if(a[i][j] == ‘T‘)
                b[3]++;
            }
            for(int i = 1; i < 4; i ++)
               if(b[i] > b[maxum])
                   maxum = i;
             printf("%c",c[maxum]);
             for(int i = 0; i < 4; i ++)
             if(i != maxum )
                hamming += b[i];
           maxum = 0;
         }
         printf("\n%d\n",hamming);
         hamming = 0;
    }
    return 0;
}