【杭电OJ1002】A + B Problem II

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最新推荐文章于 2021-11-21 13:49:51 发布

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分类专栏： c语言文章标签：杭电OJ

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本文详细解析了A+B Problem II的算法实现，介绍了如何处理大整数加法问题，包括字符串转换、大数运算和结果输出等关键步骤，提供了一个完整的C++代码示例。

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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): 488297 Accepted Submission(s): 94234

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<cstdio> 
#include<iostream>
#include<cstring>
using namespace std;
int main(){ 
	char a[1001],b[1001];	
	int lena,lenb,n;
	int na[1001],nb[1001],result[1001];
	cin>>n;
	for(int i=1;i<=n;i++) {
        /*杭电题 需要初始化数组*/
		memset(result,0,sizeof(result));
        memset(na,0,sizeof(na));
        memset(nb,0,sizeof(nb));
		scanf("%s %s",a,b);
		lena = strlen(a);
		lenb = strlen(b);
		//反转字符串 并转为为数字数组 
		for(int j=0;j<lena;j++){
			na[lena-1-j] = a[j]-'0';
		}
		for(int j=0;j<lenb;j++){
			nb[lenb-1-j] = b[j]-'0';
		} 
		int pre =0;
		int lenx = lena>lenb?lena:lenb;
		for(int j=0;j<lenx;j++){
			result[j] = na[j]+nb[j]+pre/10;
			pre=result[j];
		} 
		if(pre>9){
			result[lenx] = pre/10%10;
			lenx++;
		}
		printf("Case %d:\n",i);
        printf("%s + %s = ",a,b);
        for(int j=lenx-1;j>=0;j--)
            printf("%d",result[j]%10);
        printf("\n"); 
        if(i!=n)
            printf("\n");
	} 
	return 0; 
}