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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本文介绍了一个经典的编程问题“A+B Problem II”，该问题要求计算两个大整数的和。文章详细展示了使用 C 语言解决此问题的过程及注意事项，包括输入输出格式、字符串到整数的转换等。

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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): 4112 Accepted Submission(s): 1434

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

今天很懵，一开始Java疯狂编译错误，WA，后来发现长度是1000，改用c来做；继续疯狂WA,这个题目这样都wa，我怕是傻子哟，最后发现一个参数弄错了。。。chiou；

除了中间的ac，其他都是我（微笑）

代码如下：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char a[1005];
char b[1005];
int c[1005];
int main()
{
	int t,lena,lenb,lenc;
	cin>>t;
	for(int i=1;i<=t;i++)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		int j;
		scanf("%s%s",a,b);
		lena=strlen(a)-1;
		lenb=strlen(b)-1;
		int index=0;
		while(lena>=0&&lenb>=0)
		{
			c[index]=(c[index]+a[lena]+b[lenb]-'0'-'0');
			c[index+1]=c[index]/10;
			c[index]=c[index]%10;
			lena--;
			lenb--;
			index++;
		}
		while(lena>=0)
		{
			c[index++]+=a[lena--]-'0';
		}
		while(lenb>=0)
		{
			c[index++]+=b[lenb--]-'0';
		}
		printf("Case %d:\n",i);
		printf("%s + %s = ",a,b);
		for(int l=index;l>=0;l--)
		{
			if(c[l]==0&&l==index)
			continue;
			printf("%d",c[l]);
		}
		printf("\n");
		if(i<t)
		printf("\n");
	}
	return 0;
}