Problem 1017 - Duan's problem

/*

Time Limit: 1000MS   Memory Limit: 65536KB   Difficulty:
Total Submit:
179  Accepted: 42  Special Judge: No

Duan was a naulty boy. One day, he
made a mistake again. His teacher was very angry, and want to punish him. The
teacher wanted him to solve a problem. If he successfully solved it, he could go
home.

Here is the problem. Duan is given two number a and b. As
everybody knows, a number is made of 0-9. The teacher wanted him to count how
many 0-9 in a*b, and calculate which one appears more than others.

For
example, if a is 100, b is 4, then a*b is 400.
Number 0 1 2 3 4 5 6 7 8 9

Frequency 2 0 0 0 1 0 0 0 0 0
so duan should answer 0.
If a is 110,b
is 10,then a*b is 1100
Number 0 1 2 3 4 5 6 7 8 9
Frequency 2 2 0 0 0 0
0 0 0 0
so duan should answer 0 and 1.

Can you help
him?

Input

There are multiple test cases.Each case contains two
integers a and b (0 <= a < 10^10000, 0<=b<10^9). Input file ends of
EOF.

Output

On a single line, print the numbers of largest
frequency in incremental order separated by one space.

Sample
Input

100 4
110 10

Sample Output

0
0 1

*/

也算是大数乘法，这里a的范围是10^10000，所以用数组存，b可用int存下，这里模拟手算乘法。下面是AC代码。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char s[10020];
int b,len,ans[10];
int main()
{
    int i,f,j,max;
    while(scanf("%s%d",s,&b)!=EOF)
    {
        memset(ans,0,sizeof(ans));
        f=0;
        max=0;
        len=strlen(s);
        while(len--)
        {
            f+=(long long)b*(s[len]-'0');
            ans[f%10]++;
            f/=10;
        }
        while(f)
        {
            ans[f%10]++;
            f/=10;
        }
        for(i=0;i<10;i++)
        {
            if(max<ans[i]) {max=ans[i];j=i;}
        }
        printf("%d",j);
        for(i=j+1;i<10;i++)
        {
            if(ans[i]==max) printf(" %d",i);
        }
        printf("\n");
    }
    return 0;
}