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： hdu ACM 水题128题文章标签： integer output each input

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本文介绍了一个经典的大数加法问题“A+B Problem II”，详细解释了如何通过编程解决两个非常大的整数相加的问题，并提供了完整的C语言实现代码。

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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): 125299 Accepted Submission(s): 24132

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

Author

Ignatius.L

分析：大数相加问题

代码：

#include<stdio.h>
#include<string.h>
#define MAX 1200

int main()
{
    int n;
    scanf("%d",&n);
    int sum[MAX];
    char num1[MAX];
    char num2[MAX];
    int count=0;
    getchar();
    while(n--)
    {
        count++;
        memset(sum,0,sizeof(sum)); 
        int len1,len2,i,len;
        scanf("%s",num1);
        scanf("%s",num2);
        len1=strlen(num1);
        len2=strlen(num2);
        for(i=len1-1;i>=0;i--)
            sum[len1-i-1]=num1[i]-48;
        for(i=len2-1;i>=0;i--)
            sum[len2-i-1]+=num2[i]-48;
        int flag=0;
        for(i=0;i<MAX;i++)
        {
            int temp=flag+sum[i];
            flag=temp/10;
            sum[i]=temp%10;
        }
        i=MAX-1;
        while(sum[i]==0)
        {
            i--;
        }
        if(i<0)
            len=i+1;
        else
            len=i;
        printf("Case %d:\n",count);
            printf("%s",num1);
        printf(" + ");
            printf("%s",num2);
        printf(" = ");
        for(i=len;i>=0;i--)
            printf("%d",sum[i]);
        printf("\n");
        if(n!=0)
        printf("\n");
    }
    return 0;
}