NYOJ 103 A+B Problem II

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

AC码：

#include
#include
#include
int main()
{
	int T,i,k,len[2];
	char str[1010],ch[1010];
	int num[2][1010],max,count=0;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%s",str);
		scanf("%s",ch);
		len[0]=strlen(str);
		len[1]=strlen(ch);
		max=len[0]>len[1]?len[0]:len[1];
		memset(num,0,sizeof(num));
		for(i=0;str[i]!='\0';i++)
		{
			num[0][len[0]]=str[i]-'0';
			len[0]--;
		}
		for(i=0;ch[i]!='\0';i++)
		{
			num[1][len[1]]=ch[i]-'0';
			len[1]--;
		}
		for(i=1;i<=max;i++)
		{
			num[0][i]=num[0][i]+num[1][i];
			if(num[0][i]>=10)
			{
				num[0][i+1]+=1;
				num[0][i]=num[0][i]%10;
			}
		}
		printf("Case %d:\n",++count);
		printf("%s + %s = ",str,ch);
		if(num[0][max+1]!=0)
			printf("%d",num[0][max+1]);
		for(i=max;i>0;i--)
			printf("%d",num[0][i]);
		printf("\n");
	}
	return 0;
}