hdu 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 261958    Accepted Submission(s): 50702

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

 

题意：两个大数相加

分析：用数组模拟

//现在看，当初写的代码好幼稚，写的这么繁琐
#include<iostream>
#include<cstring>
using namespace std;
int main(){
    int T,cot=1;
    char a[1005],b[1005],c[1005],t[1005];
    cin>>T;
    while(cot<=T){
        cout<<"Case "<<cot<<":
";
        cin>>a>>b;
        cout<<a<<" + "<<b<<" = ";
        int la=strlen(a),lb=strlen(b);
        if(la<lb){//将a的长度变成大于b的长度的两个数
            for(int i=0;i<la;i++)
                t[i]=a[i];
            for(int i=0;i<lb;i++)
                a[i]=b[i];
            for(int i=0;i<la;i++)
                b[i]=t[i];
            int temp=la;
            la=lb;
            lb=temp;
        }
        if(la>lb) {//补零
            int i,j=lb-1;
            for(i=la-1;i>=la-lb;i--)
                b[i]=b[j--];
            for(;i>=0;i--)
                b[i]='0';
        }
        int flag=0;
        for(int i=la;i>=1;i--){//相加
            c[i]=(a[i-1]-'0'+b[i-1]-'0'+flag)%10+'0';
            flag=(a[i-1]-'0'+b[i-1]-'0'+flag)/10;
        }
        if(flag){//是1补一输出，否则直接输出
            c[0]='1';
            for(int i=0;i<=la;i++)
                cout<<c[i];
            cout<<endl;
        }
        else{
            for(int i=1;i<=la;i++)
                cout<<c[i];
            cout<<endl;
        }
        if(cot!=T) cout<<endl;
        cot++;
    }
    return 0;
}