大数除法。。。赶得好急,借鉴牛人的写法了

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MaxLen 200
int SubStract(int *p1,int *p2,int len1,int len2)
{
    int i;
    if(len1<len2)
         return -1;
    if(len1==len2)
    {
        for(i=len1-1;i>=0;i--)
        {
            if(p1[i]>p2[i])
                break;
            else if(p1[i]<p2[i])
                return -1;
        }
    }
    for(i=0;i<=len1-1;i++)
    {
        p1[i]-=p2[i];
        if(p1[i]<0)
        {
            p1[i]+=10;
            p1[i+1]--;
        }
    }
    for(i=len1-1;i>=0;i--)
        if(p1[i])
           return (i+1);
    return 0;
}

int main()
{
    int n,k,i,j;
    int len1,len2;
    int nTimes;
    int nTemp;
    int num_a[MaxLen];
    int num_b[MaxLen];
    int num_c[MaxLen];
    char str1[MaxLen+1];
    char str2[MaxLen+1];
   
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",str1);
        scanf("%s",str2);
       
        memset(num_a,0,sizeof(num_a));
        memset(num_b,0,sizeof(num_b));
        memset(num_c,0,sizeof(num_c));
       
        len1=strlen(str1);
        len2=strlen(str2);
       
        for(j=0,i=len1-1;i>=0;j++,i--)
           num_a[j]=str1[i]-'0';
        for(j=0,i=len2-1;i>=0;j++,i--)
           num_b[j]=str2[i]-'0';
          
        if(len1<len2)
        {
            printf("0 ");
            continue;
        }
        nTimes=len1-len2;
        for(i=len1-1;i>=0;i--)
        {
            if(i>=nTimes)
               num_b[i]=num_b[i-nTimes];
            else
               num_b[i]=0;
        }
        len2=len1;
        for(j=0;j<=nTimes;j++)
        {
            while((nTemp=SubStract(num_a,num_b+j,len1,len2-j))>=0)
            {
                len1=nTemp;
                num_c[nTimes-j]++;
            }
        }

        for(i=nTimes;num_c[i]==0&&i>=0;i--);   //原来是i=MaxLen-1,感觉没必要-.-
        if(i>=0)
            for(;i>=0;i--)
                printf("%d",num_c[i]);
        else
            printf("0");
        printf(" ");
    }
    return 0;
}
      

原文地址:https://www.cnblogs.com/get-an-AC-everyday/p/4169675.html