Hdu oj 1002 A + B Problem II（大数加法）

最新推荐文章于 2024-05-09 00:45:37 发布

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最新推荐文章于 2024-05-09 00:45:37 发布

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本文介绍了一种解决大数加法问题的方法，通过模拟手动计算的过程来处理长度可能超过1000位的大整数相加。文章提供了一个具体的编程实现案例，包括输入输出格式、算法思路及代码实现。

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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): 308383 Accepted Submission(s): 59612

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
  
  
  
  
   
    
  
  
  
  
   
   大数加法，模拟手算，注意输出格式
  
  
  
  
   
   #include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int maxn = 1005;
char a[maxn],b[maxn];
int c[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    int ans = 0;
    while(t--)
    {
        getchar();
        ans++;
        scanf("%s%s",&a,&b);
        int len_a = strlen(a);
        int len_b = strlen(b);
        int temp = 0;
        int cnt = 0;
        if(len_a >= len_b)
        {
            for(int i=len_b-1;i>=0;i--)
            {
                c[cnt] = ((a[len_a - cnt - 1] - '0') + (b[i] - '0') + temp) % 10;
                temp = ((a[len_a - cnt - 1] - '0') + (b[i] - '0') + temp) / 10;
                cnt++;
            }
            for(int i=len_a-len_b-1;i>=0;i--)
            {

                c[cnt] = ((a[i] - '0')+ temp) % 10;
                temp = ((a[i] - '0') + temp) / 10;
                cnt++;
            }
        }
        else
        {
            for(int i=len_a-1;i>=0;i--)
            {
                c[cnt] = ((b[len_b - cnt - 1] - '0') + (a[i] - '0') + temp) % 10;
                temp = ((b[len_b - cnt - 1] - '0') + (a[i] - '0') + temp) / 10;
                cnt++;
            }
            for(int i=len_b-len_a-1;i>=0;i--)
            {

                c[cnt] = ((b[i] - '0')+ temp) % 10;
                temp = ((b[i] - '0') + temp) / 10;
                cnt++;
            }
        }
        printf("Case %d:\n",ans);
        printf("%s + %s = ",a,b);
        if(temp)
            printf("%d",temp);
        for(int i=cnt-1;i>=0;i--)
        {
            printf("%d",c[i]);
        }
        printf("\n");
        if(t)
            printf("\n");
    }
}