一个二维数组中寻找字符串出现的次数，例如
Mtauy
Eedio
Adada
Gbdac
Crawm       其中ada出现了三次，
从左往右，从右往左，从上往下，从下往上读
可以部分重叠，但不可以全部重叠
如何编程

不行。。。。显示0个，这个程序我也快做好了，就差如果是回文的时候要除以2了。。。。。
#include<stdio.h>
#include<string.h>
int main()
{
    char str[110][110],s[110];
    int m,n,i,j,k,count=0,a,b;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
    for(i=0;i<n;i++)
    {
        for(j=0;j<=m;j++)
        {
            scanf("%c",&str[i][j]);
        }
        if(i==(n-1)&&j==m)
        {
            str[i][j]='\0';
        }
    }
    getchar();
    gets(s);
    for(a=0;s[a]!='\0';a++)
    {}
    for(i=0;i<n;i++)//1
    {
        for(j=0;j<m;j++)
        {
            if(str[i][j]==s[0])
            {
                for(k=0;;k++)
                {
                    if(str[i][j+k]==s[k])
                    {
                        continue;
                    }
                    if(s[k]==s[a])
                    {
                        count++;
                    }
                    if(str[i][j+k]!=s[k])
                    {
                        break;
                    }

                }
            }
        }
    }printf("1:%d\n",count);
    for(i=0;i<m;i++)//2
    {
        for(j=0;j<n;j++)
        {
            if(str[j][i]==s[0])
            {
                for(k=0;;k++)
                {
                    if(str[j+k][i]==s[k])
                    {
                        continue;
                    }
                    if(s[k]==s[a])
                    {
                        count++;
                    }
                    if(str[j+k][i]!=s[k])
                    {
                        break;
                    }
                }
            }
        }
    }printf("2:%d\n",count);
    for(i=n;i>=0;i--)//3
    {
        for(j=m;j>=0;j--)
        {
            if(str[i][j]==s[0])
            {
                for(k=0;;k++)
                {
                    if(str[i][j-k]==s[k])
                    {
                        continue;
                    }
                    if(s[k]==s[a])
                    {
                        count++;
                    }
                    if(str[i][j-k]!=s[k])
                    {
                        break;
                    }
                }
            }
        }
    }printf("3:%d\n",count);
    for(i=m;i>=0;i--)//4
    {
        for(j=n;j>=0;j--)
        {
            if(str[j][i]==s[0])
            {
                for(k=0;;k++)
                {
                    if(str[j-k][i]==s[k])
                    {
                        continue;
                    }
                    if(s[a]==s[k])
                    {
                        count++;
                    }
                    if(str[j-k][i]!=s[k])
                    {
                        break;
                    }
                }
            }
        }
    }printf("4:%d\n",count);
     printf("%d\n",count);
    count=0;
    }
    return 0;
}