87分，过不去最后一个测试点，求解

Yideer2 @ 2024-11-23 11:29:00

#include <stdio.h>
#include<string.h>
char l1[10005];
char b[10005];
char c[10005];
long long int a;int x;
//b/a=c
int main()
{
    scanf("%s",l1);
    scanf("%lld",&a);
    if(l1[0]=='0') printf("0");
    //printf("%lld %s",a,l1); 
    int i;
    int lb=strlen(l1);
    for(i=1;i<=lb;i++){
        b[i]=l1[i-1]-'0';
    }
    for(i=1;i<=lb;i++){
        c[i]=(x*10+b[i])/a;
        x=(x*10+b[i])%a;
        //printf("%ld ",c[i]);
    }
     int lc=1;
    for(lc=1;c[lc]==0;lc++);
    //printf("%d ",lc);
    for(i=lc;i<=lb;i++){
        printf("%d",c[i]);
    }
    return 0;
}

by leo_liaocr @ 2024-11-23 11:36:41

@Yideer2 x的类型改long long

by Yideer2 @ 2024-11-23 11:45:32

@leo_liaocr 还是不行

by leo_liaocr @ 2024-11-23 11:47:09

@Yideer2 c数组改int

by Yideer2 @ 2024-11-23 11:48:54

@leo_liaocr 哎，还是不行

by leo_liaocr @ 2024-11-23 12:23:55

@Yideer2 ? 我试了可以啊 你看看

#include <stdio.h>
#include<string.h>
char l1[10005];
char b[10005];
int c[10005];
long long int a;long long x;
//b/a=c
int main()
{
    scanf("%s",l1);
    scanf("%lld",&a);
    if(l1[0]=='0') printf("0");
    //printf("%lld %s",a,l1); 
    int i;
    int lb=strlen(l1);
    for(i=1;i<=lb;i++){
        b[i]=l1[i-1]-'0';
    }
    for(i=1;i<=lb;i++){
        c[i]=(x*10+b[i])/a;
        x=(x*10+b[i])%a;
        //printf("%ld ",c[i]);
    }
     int lc=1;
    for(lc=1;c[lc]==0;lc++);
    //printf("%d ",lc);
    for(i=lc;i<=lb;i++){
        printf("%d",c[i]);
    }
    return 0;
}

by Yideer2 @ 2024-11-23 12:27:17

@leo_liaocr我的问题，确实可以，谢谢了