UVA 10106 Product

将数学运算过程程序化，注意地址的变动，缩位进位就可以啦！

#include<stdio.h>
#include<string.h>
#define MAXN1 250 + 10
#define MAXN2 500 + 10
char s1[MAXN1], s2[MAXN1];
int num[MAXN2],num1[MAXN1],num2[MAXN1];
void solve()
{
    int len1 = strlen(s1);
    for(int i = 0; i < len1; i ++)
        num1[i] = s1[len1-i-1] - '0';
    int len2 = strlen(s2) ;
    for(int j = 0; j < len2; j ++)
        num2[j] = s2[len2-1-j] - '0';
            for(int p = 0; p < len2; p ++)
            {
                for(int q = 0; q < len1; q ++)
                {
                    num[p+q] += num2[p] * num1[q];//printf("num[p+q]=%d",num[p+q]);
                    if(num[p+q]>9)
                    {
                        num[p+q+1] += num[p+q]/10;
                        num[p+q] %= 10;
                    }
                }
            }
}
void output()
{
    int i;
    for(i = MAXN2; i >= 0; i --)
    {
        if(i == 0) break;
        if(num[i] != 0) break;
    }
    for(int j = i; j >=0; j --)
    printf("%d",num[j]);
    printf("\n");
}
void input()
{
    while(scanf("%s",s1) == 1)
    {
        memset(num,0,sizeof(num));
        scanf("%s",s2);
        solve();
        output();
    }
}
int main()
{
    input();
    return 0;
}