/*
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209912 Accepted Submission(s): 40404
Problem Description
2 1 2 112233445566778899 998877665544332211
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
*/
#include<stdio.h>
#include<string.h>
#define N 1000
int main(){
int n;
while(~scanf("%d",&n)){
//吞掉回车符
getchar();
int i,j,k,a,b,t;
//数组定义
char schar1[N+10]={‘\0‘};
char schar2[N+10]={‘\0‘};
int sint1[N+10];
int sint2[N+10];
for(k=1;k<=n;k++){
memset(sint1,0,sizeof(sint1));
memset(sint2,0,sizeof(sint2));
scanf("%s %s",schar1,schar2);
a=strlen(schar1);
b=strlen(schar2);
//反转赋值
for(i=0; i<a; i++)
sint1[a-1-i]=schar1[i]-‘0‘;
for(i=0; i<b; i++)
sint2[b-1-i]=schar2[i]-‘0‘;
//将a定义为a,b中最大的
if(a<b){
t=a;
a=b;
b=t;
}
//判断是否进位
for(i=0;i<=a;i++){
sint1[i]+=sint2[i];
if(sint1[i]>=10){
sint1[i]-=10;
sint1[i+1]++;
}
}
printf("Case %d:\n%s + %s = ",k,schar1,schar2);
//判断是否有进位,不同情况不同输出
if(sint1[a]==0){
for(i=a-1;i>=0;i--)
printf("%d",sint1[i]);
printf("\n");
}
if(sint1[a]!=0){
for(i=a;i>=0;i--)
printf("%d",sint1[i]);
printf("\n");
}
if(k!=n) printf("\n"); //最后一个不换行
}
}
return 0;
}
大数A+B 【杭电-1002】 附题,布布扣,bubuko.com
原文地址:http://blog.csdn.net/holyang_1013197377/article/details/38302757