杭电1002 大数相加

发布时间：2020-12-14 03:30:55 所属栏目：大数据来源：网络整理

导读：http://acm.hdu.edu.cn/showproblem.php?pid=1002 A + B Problem II 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

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

A + B Problem II

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

?第一次算大数相加，在网上看了思想，自己敲代码，由于粗心数组初始化出了点问题，提交了n次都没过，还好细心的队友发现啦！

#include<iostream>
#include<cstdio>
#include<string.h>
#define N 1005
int main()
{
	int i,j,k,m,l1,l2,t;
	char s1[N],s2[N];
	int a[N],b[N];
	scanf("%d",&t);
	for(m=1;m<=t;m++)
	{
		memset(s1,sizeof(s1));
		memset(s2,sizeof(s2));
		memset(a,sizeof(a));
		memset(b,sizeof(b));
		scanf("%s%s",&s1,&s2);
		l1=strlen(s1);
		l2=strlen(s2);
		if(l1>l2)
			k=l1;
		else
			k=l2;
		for(i=k,j=l1-1;j>=0;i--,j--)
			a[i]=s1[j]-'0';
		for(i=k,j=l2-1;j>=0;i--,j--)
			b[i]=s2[j]-'0';
		for(i=k;i>0;i--)
		{
			a[i]+=b[i];
			if(a[i]>=10)
			{
				a[i]=a[i]%10;
				a[i-1]++;
			}
		}
		printf("Case %d:n%s + %s = ",s1,s2);
		if(a[0]!=0)
		{
			for(i=0;i<=k;i++)
			{
				printf("%d",a[i]);
			}
		}
		else
		{
			for(i=1;i<=k;i++)
				printf("%d",a[i]);
		}
		if(m<=t-1)
			printf("nn");
		else
			printf("n");
	}
	return 0;
}

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!