输出
                            *                     
                            *                     
        *                   *                     
        *                   *     *   *            
        *                   *     *   *            
*       *     *             *     *   *            
*       *     * *     * *   *     * * *            
*       *   * * *     * *   * *   * * * *         
*     * * * * * *     * * * * *   * * * *     * *  
* * * * * * * * * * * * * * * * * * * * * * * * * *
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

分别输出A—Z上*的数目

就是求代码 唉唉

输入4行全部由大写字母组成的文本，输出一个垂直直方图，给出每个大写字母出现的次数。
注意：只用输出大写字母的出现次数，不用输出空白字符，数字或者标点符号的出现输出次数。

#include<stdio.h>
#define N 1000
main()
{
    char a[N],b[N],c[N],d[N];
    int  i,j,count[N],counta[N],countb[N],countc[N],countd[N],max = 0;
    gets(a);
    gets(b);
    gets(c);
    gets(d);
    for(i = 0;i < 26;i++)
    {
        counta[i] = 0;
        countb[i] = 0;
        countc[i] = 0;
        countd[i] = 0;
    }
    for(i = 0;i < 26;i++)
    {
        for(j = 0;a[j] != '\0';j++)
        {
            if(a[j] == i+65)
                counta[i]++;
        }
        for(j = 0;b[j] != '\0';j++)
        {
            if(b[j] == i+65)
                countb[i]++;
        }
        for(j = 0;c[j] != '\0';j++)
        {
            if(c[j] == i+65)
                countc[i]++;
        }
        for(j = 0;d[j] != '\0';j++)
        {
            if(d[j] == i+65)
                countd[i]++;
        }
        count[i] = counta[i]+countb[i]+countc[i]+countd[i];
        if (count[i] > max)
            max = count[i];
    }
    for ( i = 0; i < max;i++)
    {
        for(j = 0;j < 26;j++)
        {
            if(max - count[j] - i > 0)
                printf("  ");
            else
                printf("* ");
        }
        printf("\n");
    }
    for(i = 0;i < 26;i++)
    {
        printf("%c ",i+65);
    }
    printf("\n");
}

已经解决了 呵呵