A+B Problem II

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

//最开始写的较笨的方法

#include<stdio.h>
#include<string.h>
int main()
{
	int n,x,y,i,j,d,p;
	char a[1000],b[1000],e[1001];
	scanf("%d",&n);
	for(d=1;d<=n;d++)
	{
		scanf("%s %s",&a,&b);
		printf("Case %d:
",d);
		x=strlen(a)-1;
		y=strlen(b)-1;
		p=0;
		for(i=0;x>=0||y>=0;x--,y--)
		{
			if(x>=0&&y>=0)
				e[i]=a[x]+b[y]-'0'+p;
			if(x>=0&&y<0)
				e[i]=a[x]+p;
			if(x<0&&y>=0)
				e[i]=b[y]+p;
			p=0;
			if(e[i]>'9')
			{
				e[i]=e[i]-10;
				p=1;
			}
			i++;
		}
		printf("%s + %s = ",a,b);
		if(p==1)
			printf("1");
		for(j=i-1;j>=0;j--)
			printf("%d",e[j]-'0');   //printf("%c",e[j]);
		printf("
");
		if(d!=n)
			printf("
");
	}
	return 0;
}

//之后的方法

<span style="font-size:18px;">#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	char a[1000],c[1000];	
	int i,j,x,n,d;
	cin>>n;
    for(d=1;d<=n;d++)
	{	
		int sum[1000]={0};
		cin>>a>>c;
		cout<<"Case "<<d<<":"<<endl;
		cout<<a<<" + "<<c<<" = ";
		int t1=strlen(a);
		int t2=strlen(c);
		x=999;
		for(i=t1-1,j=t2-1;i>=0&&j>=0;i--,j--)
			sum[x--]=(a[i]-'0')+(c[j]-'0');
		if(i>=0)
			sum[x--]=a[i--]-'0';
		if(j>=0)
			sum[x--]=c[j--]-'0';
		x=0;
		for(i=999;i>=0;i--)
		{
			sum[i]+=x;
			x=0;
			if(sum[i]>9)
			{
				x=sum[i]/10;
				sum[i]=sum[i]%10;
			}
		}
	    for(i=0;i<1000;i++)
		{
			if(sum[i]!=0)
				break;
		}
		for(j=i;j<1000;j++)
			cout<<sum[j];
		cout<<endl;
		if(d!=n)
			cout<<endl;
	}
	return 0;
}</span>