hdoj1002(大数加法)

最新推荐文章于 2019-04-18 21:43:47 发布

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本文介绍了一个经典的编程竞赛题目“A+B Problem II”的详细解析。该题目要求计算两个非常大的整数相加的结果，并提供了具体的输入输出样例。文章还分享了一段使用 C 语言实现的解题代码，通过逐位相加的方法来处理大整数加法的问题。

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60420 Accepted Submission(s): 11074

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

#include<stdio.h>
#include<string.h>
char str1[1005],str2[1005];
int num1[1005],num2[1005];
int main()
{
    int n,l=1,m=1;
    scanf("%d",&n);
    while(n--)
    {
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        scanf("%s%s",str1,str2);
        int len1=strlen(str1);
        int len2=strlen(str2);
        int i,j=0;
        for(i=len1-1;i>=0;i--)
        num1[j++]=str1[i]-'0';
        j=0;
        for(i=len2-1;i>=0;i--)
        num2[j++]=str2[i]-'0';
        for(i=0;i<1005;i++)
        {
            num2[i]=num2[i]+num1[i];
            if(num2[i]>=10)
            { num2[i]=num2[i]-10;
              num2[i+1]++;
            }
        }
        if(m==1) m=2;
        else printf("/n");
        printf("Case %d:/n",l++);
        printf("%s + %s = ",str1,str2);
        bool bStartOutput=false;
        for(i=1004;i>=0;i--)
        {
            if(num2[i]||bStartOutput)
            {printf("%d",num2[i]);
             bStartOutput=true;
            }
        }
        if(!bStartOutput) printf("0");
        printf("/n");
    }
    return 0;
}