A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 388690    Accepted Submission(s): 75226

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

 

 

这题用大数加法做才行，或者用Java做。

发现自己好粗心啊。

#include <iostream>
#include <bits/stdc++.h>
#include <cstring>
#include <cstdio>
#include <string>
#define N 1010
using namespace std;
int n;
int k[N];
void bigdate(string a,int alen,string b,int blen){
    reverse(a.begin(),a.end());
    reverse(b.begin(),b.end());
    for(int i=0;i<alen;i++){
        int an=a[i]-'0',bn=b[i]-'0';
        if(an+bn+k[i]>9){
            k[i]=(an+bn+k[i]-10);
            k[i+1]++;
        }else{
            k[i]+=(an+bn);
        }
    }
    if(alen<blen)
        for(int i=alen;i<blen;i++){
            k[i]+=(b[i]-'0');
        }
}
int main(){
    cin>>n;
    for(int p=1;p<=n;p++){
        string a,b;
        cin>>a>>b;
        memset(k,0,sizeof(k));
        int alen=a.length();
        int blen=b.length();
        if(alen<blen)
            bigdate(a,alen,b,blen);
        else
            bigdate(b,blen,a,alen);
        cout<<"Case "<<p<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        if(k[max(alen,blen)]!=0)
            cout<<k[max(alen,blen)];
        for(int i=max(alen,blen)-1;i>=0;i--)
            cout<<k[i];
        if(p!=n)
            cout<<endl<<endl;
        else
            cout<<endl;
    }
    return 0;
}

Java代码

import java.util.*;
import java.math.*;
public class Main{
    public static void main(String[] args){
        Scanner scanner =new Scanner(System.in);
        int n=scanner.nextInt();
        for(int i=1;i<=n;i++){
            if(i!=1)
                System.out.println();
            BigInteger a,b;
            a=scanner.nextBigInteger();
            b=scanner.nextBigInteger();
            System.out.println("Case "+i+":");
            System.out.println(a+" + "+b+" = "+a.add(b));
        }
    }
}