九度OJ 1089 数字反转

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char a[10];
char b[10];
char c[10];
int a1,b1,s1,a2,b2,s2;
void reverse(char *a)
{
    int len=strlen(a);
    for(int i=0,j=len-1;i<j;++i,--j)
    {
        int t=a[i];
        a[i]=a[j];
        a[j]=t;
    }
}
int main(int argc, char *argv[])
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;++i)
        {
            scanf("%s%s",a,b);
            a1=atoi(a);
            b1=atoi(b);
            s1=a1+b1;//和
            reverse(a);
            reverse(b);
            a2=atoi(a);
            b2=atoi(b);
            s2=a2+b2;//反转的和
            int j=0;
            while(s1)
            {
                c[j++]=s1%10+'0';
                s1/=10;
            }
            c[j]='\0';//和的反转
            s1=atoi(c);
            if(s1==s2)
            {
                reverse(c);
                printf("%s\n",c);
            }
            else
                printf("NO\n");
        }
 
    }
    return 0;
}
 
/**************************************************************
    Problem: 1089
    User: kirchhoff
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:912 kb
****************************************************************/