HDOJ 1002 高精度加法

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 290083 Accepted Submission(s): 55742

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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
//
//  main.cpp
//  1003
//
//  Created by 张嘉韬 on 16/1/8.
//  Copyright © 2016年 张嘉韬. All rights reserved.
//
#include <iostream>
#include <cstring>
using namespace std;
int main(int argc, const char * argv[]) {
    //freopen("/Users/zhangjiatao/Desktop/input.txt","r",stdin);
    char a1[2000],b1[2000];
    int a[2000],b[2000],c[2000],lena,lenb,lenc,temp;
    int n;
    cin>>n;
    for(int t=1;t<=n;t++)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        cin>>a1>>b1;
        lena=strlen(a1);
        lenb=strlen(b1);
        for(int i=0;i<lena;i++) a[lena-i]=a1[i]-48;
        for(int i=0;i<lenb;i++) b[lenb-i]=b1[i]-48;
        lenc=1;
        while(lenc<=lena||lenc<=lenb)
        {
            temp=a[lenc]+b[lenc]+c[lenc];
            if(temp>=10) c[lenc]=temp-10,c[lenc+1]++;
            else  c[lenc]=temp;
            lenc++;
        }
        if(c[lenc]==0) lenc--;
        cout<<"Case "<<t<<":"<<endl;
        //cout<<lena<<" "<<lenb<<endl;
        for(int i=lena;i>=1;i--) cout<<a[i];
        cout<<" + ";
        for(int i=lenb;i>=1;i--) cout<<b[i];
        cout<<" = ";
        for(int i=lenc;i>=1;i--) cout<<c[i];
        cout<<endl;
        if(t!=n) cout<<endl;
    }
    return 0;
}
总结：
1.当题目解决后应该及时的进行总结，如果一段时间之后再总结，当时的一些重要的经验和教训很难被积累下来，这样的话效率比较低。
2.可以用字符串来储存特别大的正数，输入方法也很简单直接用cin就行，获取长度用strlen，注意这些方法都试使用字符数组的时候用的，如果要使用string的话，获取长度要用name.size()或name.length()(两个函数在功能上没有区别)

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