大数相加

1199 - A+BProblemII

时间限制：3秒 内存限制：128兆

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题目描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入

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.

输出

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.

样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
	int t,count=1;
	cin>>t;
	while(count!=t+1)
	{
		char a[1010],b[1010],c[1010];
		scanf("%s%s",a,b);
		int i=strlen(a)-1,j=strlen(b)-1;
		int x,y,z,up=0,k=0;
		while(i>=0||j>=0)
		{
			if(i<0)x=0;
			else x=a[i]-'0';
			if(j<0)y=0;
			else y=b[j]-'0';
			z=x+y+up;
			if(z>9) 
			{up=1;z=z-10;}
			else up=0;
			c[k++]=z+'0';
			i--,j--;
		}
		if(up)
		c[k++]='1';
		c[k]='\0';
		cout<<"Case "<<count++<<":"<<endl;
		cout<<a<<" + "<<b<<" = ";
		for(i=k-1;i>=0;i--)
		cout<<c[i];cout<<endl;
	}
	
}