HDOJ 1002 A + B Problem II【大数，基础】

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最新推荐文章于 2024-02-08 13:38:17 发布

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本文介绍了一道经典的A+B问题的变种——A+B Problem II，该问题要求使用数组模拟大整数加法来解决非常大的整数相加的问题。文章提供了详细的解题思路和完整的C语言代码实现。

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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): 314078 Accepted Submission(s): 60863

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

模拟+大数。

最基础的大数题，部分细节详见代码。总的来说就是用数组模拟10进制运算，注意位数问题。

#include<stdio.h>
#include<string.h>
char a[1111],b[1111],c[1111],d[1111],sum[1111];
int main()
{
	int T,i,j;
	scanf("%d",&T);
	int n=T;
	int k=1;
	while(T--)
	{
		memset(c,0,sizeof(c));
		memset(d,0,sizeof(d));
		memset(sum,0,sizeof(sum));
		scanf("%s%s",a,b);
		int l1=strlen(a);
		int l2=strlen(b);
		for(i=l1-1,j=0;j<l1;j++,i--)
			c[j]=a[i];
		for(i=l2-1,j=0;j<l2;j++,i--)
			d[j]=b[i];
		int l=l1;
		if(l2>l)
			l=l2;
		int p=0,q=0;
		for(i=0;i<l;i++)
		{
			if(c[i]-'0'>=0)
				p+=c[i]-'0';
			if(d[i]-'0'>=0)
				p+=d[i]-'0';
			p+=q;
			q=p/10;
			p=p%10;
			sum[i]=p+'0';
			p=0;
		}
		if(q!=0)
			sum[i++]=q+'0';
		printf("Case %d:\n",k++);
		printf("%s + %s = ",a,b);
		for(j=i-1;j>=0;j--)
			printf("%c",sum[j]);
		printf("\n");
		if(k!=n+1)
			printf("\n");
	}
	return 0;
}