HDU 1002 A + B Problem II (大数加法)

发布时间：2020-12-14 02:18:44 所属栏目：大数据来源：网络整理

导读：A + B Problem II ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

A + B Problem II

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?Total Submission(s): 271671????Accepted Submission(s): 52508

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 = 2222222222222221110

? ? ? ?大数问题，就是用数组模拟计算问题。加法，用数组模拟计算数的加法，进行对每一位进行相加。

附上代码：

 
#include<iostream>
#include<algorithm>
#include<cstring>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    char a[1050],b[1050];
    int c[1050],d[1050],sum[1050];
    int n,num = 0;
    cin >> n;
    getchar();
    while(n--)
    {
        num++;
        memset(c,sizeof(c));
        memset(d,sizeof(d));
        memset(sum,sizeof(sum));
        cin >> a >> b;
        int j = 0;
        for(int i = strlen(a) - 1;i >= 0;i--)    //存入数组中 
        {
            c[j++] = a[i] - '0';
        }
        j = 0;
        for(int i = strlen(b) - 1;i >= 0;i--)     //存入数组中 
        {
            d[j++] = b[i] - '0';
        }
        int len = strlen(a) > strlen(b) ? strlen(a) : strlen(b);  //   寻找出最长的数字位数 
        int jin = 0;
        for(int i = 0;i < len;i++)
        {
            sum[i] = c[i] + d[i] + jin ;
            jin = 0;
            if(sum[i] > 9 && i < len - 1)         //  i  =  len - 1 的时候为最前面的一位，所以不需要进行进位 
            {
                jin = sum[i]/10;
                sum[i] = sum[i]%10;
            }
        }
        if(num != 1)          //  控制输出的格式 
		cout << endl;
        cout << "Case " << num << ":" << endl;
        cout << a << " + " << b << " = ";
        for(int i = len-1;i >= 0;i--)
        {
            cout << sum[i];       
        }
        cout << endl;
     }
     return 0;
}

（编辑：李大同）

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