【杭电-oj】-1002-A + B Problem II（大数相加）

发布时间：2021-03-01 18:09:54 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) Total Submission(s): 314006????Accepted Submission(s): 60840 Problem Description I have a very simple problem for you. Given two integers

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 314006????Accepted Submission(s): 60840

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.
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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.
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Sample Input

  
  
   
   2
1 2
112233445566778899 998877665544332211

?
Sample Output

  
  
   
   Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110

用字符串存比较大的数，然后把数倒着写，因为比较利于计算，此时应注意方法，再判断最高位有没有进位即可。

但是师傅建议我这么写，记下，下道题试试。

初始化：

【杭电-oj】-1002-A + B Problem II（大数相加）

输出：

【杭电-oj】-1002-A + B Problem II（大数相加）

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int n,num=0;
	char a[1010],b[1010];			//把数当做字符串用a,b存
	int l1,l2,l3;
	int numa[1010]={0},numb[1010]={0};		 //分别存储a,b准换为数字后的值 
	scanf("%d",&n);
	while(n--)
	{
		memset (numa,sizeof (numa));
		memset (numb,sizeof (numb));
		scanf("%s %s",a,b);
		l1=strlen(a);
		l2=strlen(b);
		l3=max(l1,l2);
		for(int i=0;i<l1;i++)
			numa[l1-i-1]=a[i]-'0';			//为了方便计算，把数倒着存，注意此处的技巧，会常用 
		for(int i=0;i<l2;i++)
			numb[l2-i-1]=b[i]-'0';
		numb[l3]=0;
		for(int i=0;i<l3;i++)
		{
			numb[i]=numa[i]+numb[i];
			if(numb[i]>9)
			{
				numb[i]=numb[i]-10;
				numb[i+1]++;
			}
		}
		num++;
		printf("Case %d:n",num);
		printf("%s + %s = ",b);
		if(numb[l3]==0)			//判断l3（最后一位）是否为零 
		{
			for(int i=l3-1;i>=0;i--)	//倒着相加，倒着输出 
				printf("%d",numb[i]);
			printf("n");
		}
		else
		{
			for(int i=l3;i>=0;i--)
				printf("%d",numb[i]);
			printf("n"); 
		} 
		if(n!=0)
			printf("n");
	}
	return 0;
}

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