HDU 1002 A + B Problem II（大数据）

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

 

Sample Input

2

1 2

112233445566778899 998877665544332211

 

Sample Output

Case 1: 1 + 2 = 3

Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

 

因为数字超级大，所以就算long long 也是不行的，所以用字符数组输入，然后再转化成数字，然后进行加法进位运算。

要注意PE

AC代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int t,len1,len2;
char a[1010],b[1010];
int na[1010],nb[1010],ans[1010];
int main(){
    scanf("%d",&t);
    int tt=t;
    int cnt=1;
    while(t--){
        scanf("%s%s",a,b);
        len1=strlen(a);
        len2=strlen(b);
        memset(na,0,sizeof(na));//初始化清零 
        memset(nb,0,sizeof(nb));
        memset(ans,0,sizeof(ans));
        for(int i=len1-1;i>=0;i--)//将字符转为数字 
            na[len1-1-i]=a[i]-'0';//要逆着来，为了后面加法进位 
        for(int i=len2-1;i>=0;i--)
            nb[len2-1-i]=b[i]-'0';
        int i,k=0;
        for(i=0;i<max(len1,len2);i++){
            k=na[i]+nb[i]+k;
            ans[i]=k%10; //进位 
            k=k/10;  
        } 
        if(k!=0)  
            ans[i++]=k;
        printf("Case %d:\n%s + %s = ",cnt++,a,b);
        int p=0;
        if(i==1&&ans[0]==0){ //如果不写这个，0+0就不输出了 
            printf("0\n");
            if((cnt-1)!=tt)
                printf("\n");  
            continue;
        }
        for(int j=i-1;j>=0;j--){
            if(p==0&&ans[j]==0){ //前导零，如0003+2 
                continue;
            }
            else{
                p=1;
                printf("%d",ans[j]);
            }
        }
        printf("\n");
        if((cnt-1)!=tt)
            printf("\n"); //PE了好几次，它要两个测试之间有空行，最后一组不能有空行 
    }
}

（突然想起来，在每篇博客后面写一句自己的话才是我的风格嘛）

（私はカメラですね）