NYOJ 103 A+B Problem II (大数)

A+B Problem II

时间限制： 3000 ms  |  内存限制： 65535 KB

难度： 3

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

注 - 此题为 ： NYOJ 103 A+B Problem II (大数)

        大数 a+b  模板   注意格式

已AC代码：

#include
#include
#define max(x,y) (x>y?x:y)

char ch1[1100],ch2[1100];
int a[1100],b[1100],la,lb;
int main()
{
	int T,CASE=1,i,j;
	scanf("%d",&T);
	while(T--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		
		scanf("%s%s",ch1,ch2);
		
		la=strlen(ch1);
		lb=strlen(ch2);
		
		for(i=0,j=la-1;i9)  //进位 
			{
				a[i]-=10;
				a[i+1]+=1;
			}
		}
		
		printf("Case %d:\n",CASE++);
		printf("%s + %s = ",ch1,ch2);
		
		for(i=1000;i>=0;--i)
			if(a[i]!=0)
				break;
		for(;i>=0;--i)
			printf("%d",a[i]);
		printf("\n");
	}
	return 0;
}