HDU 1002：A + B Problem II（大数相加）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 407964    Accepted Submission(s): 79058

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

基本思想：竖式运算，用字符串输入，然后每位数减去'0'再相加。

搜了很多代码都是计算输出的时候没有去掉前面的0，比如：000+1=001 。

看着别人的代码，顺着思路改了一下现在的输出是没有前面的0的，即：000+1=1 。

这两种写法都能AC，不去掉0的输出时比较简单。

还有就是注意输出的时候除最后一组数之外，每两组数之间有一个空行

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
const int maxn=1e6+10;
int a[maxn],b[maxn],c[maxn];
char num1[maxn],num2[maxn];
int main()
{
	int t;
	int k=1;
	int i;
	scanf("%d",&t);
	while(t--)
	{
		int sum;
		int flag=0;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		scanf("%s %s",num1,num2);
		int l1=strlen(num1);
		int l2=strlen(num2);
		for(i=0;i<l1;i++) a[i]=num1[l1-1-i]-'0';
		for(i=0;i<l2;i++) b[i]=num2[l2-1-i]-'0';//字符串转换成数字
		int ml=std::max(l1,l2);//找到最长的字符串长度
		for(sum=0,i=0;i<ml;i++)
		{
			c[i]=(a[i]+b[i]+sum)%10;
			sum=(a[i]+b[i]+sum)/10;
		}
		if(sum) c[ml]=1;
		printf("Case %d:\n",k++);
		printf("%s + %s = ",num1,num2);
		if(c[ml]==1) printf("1");
		for(i=ml-1;i>=0;i--)
		{
			if(c[i]!=0)
			{
				flag++;
				printf("%d",c[i]);
			}
			else if(c[i]==0&&flag!=0)
			{
				printf("0");
				flag++;
			}
			else if(c[i]==0&&flag==0) continue;
		}
		if(c[0]==0&&flag==0) printf("0");
		/*这种写法是不去掉开头的0的
		for(i=ml-1;i>=0;i--) printf("%d",c[i]);*/
		printf("\n");
		if(t) printf("\n");
	}
	return 0;
}