HDU-1002 A + B Problem II

最新推荐文章于 2020-08-17 22:11:42 发布

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最新推荐文章于 2020-08-17 22:11:42 发布

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分类专栏： ACM__大数问题

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本文详细介绍了HDU-1002A+B Problem II的题目要求及解决方法，通过字符串存储大数并进行加法运算的方式，解决了大数相加的问题。提供了完整的代码实现及运行示例。

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HDU-1002 A + B Problem II

Time Limit : 2000/1000ms (Java/Other)
Memory Limit : 65536/32768K (Java/Other)

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

题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1002

分析

题意：A+B大数问题，求A+B的值

思路：看数据就明白了，这是一个大数问题。大数的思想实际上就是将运算的过程实现出来。用两个字符串分别储存两个数，再转换成数组进行加法运算就可以了。大数的加法相对来说还是很容易理解的。代码很容易就可以看懂。

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int main()
{
    char num1[1005],num2[1005];
    int Bnum1[1005],Bnum2[1005];
    int n;
    cin>>n;
    for (int i=0;i<n;i++)
    {
        memset(Bnum1,0,sizeof(Bnum1));
        memset(Bnum2,0,sizeof(Bnum2));
        cin>>num1>>num2;
        int len1=strlen(num1);
        int len2=strlen(num2);
        for (int i=len1-1;i>=0;i--)
            Bnum1[len1-1-i]=num1[i]-'0';
        for (int i=len2-1;i>=0;i--)
            Bnum2[len2-1-i]=num2[i]-'0';
        int len=max(len1,len2);
        int carry=0;
        for (int i=0;i<len;i++)
        {
            Bnum2[i]+=Bnum1[i]+carry;
            carry=Bnum2[i]/10;
            if (Bnum2[i]>=10)
                Bnum2[i]-=10;
        }
        Bnum2[len]=carry;
        if (carry!=0)
            len++;
        cout<<"Case "<<i+1<<":"<<endl;
        cout<<num1<<" + "<<num2<<" = ";
        for (int i=len-1;i>=0;i--)
            cout<<Bnum2[i];
        cout<<endl;
        if (i!=n-1)
            cout<<endl;
    }
    return 0;
}