HD1002 A + B Problem II（大数简单的加法）

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本文详细介绍了如何解决A+B Problem II问题，包括输入输出规范、算法实现及代码示例。

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                      <span style="font-size:24px;"> A + B Problem II</span>

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260769 Accepted Submission(s): 50427
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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.

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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

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
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AC代码
</pre><br /></div><div class="panel_bottom"><pre class="cpp" name="code">#include<stdio.h>
#include<string.h>
#define max 1000
int a[max+10];
int b[max+10];
char str1[max+10];
char str2[max+10];
int main()
{
	int k,i,j,p=1;
	scanf("%d",&k);
	while(k--)
	{
		scanf("%s %s",str1,str2);
		memset(a,0,sizeof(a));//a数组中的元素每个都要清零。 
		memset(b,0,sizeof(b));// b数组中的元素每个都要清零。
		int m=strlen(str1);
		int n=strlen(str2);
		for(i=0,j=m-1;i<m;i++,j--)
		a[i]=str1[j]-'0';//字符串1中的各个元素倒叙转化为数字，把值赋数组a。 其中str1[j]-'0'，是把字符数字转变为整型数字。
		for(i=0,j=n-1;i<n;i++,j--)
		b[i]=str2[j]-'0';//字符串2中的各个元素倒叙转化为数字，把值赋数组b。其中str2[j]-'0'，是把字符数字转变为整型数字。 
		for(i=0;i<max;i++)
		{
			a[i]+=b[i];//逐位相加
			if(a[i]>=10)
			{             //看是否要进位
				a[i]-=10;
				a[i+1]++;
			}
		}
		for(i=max;i>=0&&(a[i]==0);i--);//把首位为零的全部清除掉。 
		if(i>=0)
		{
			printf("Case %d:\n",p++);
			printf("%s + %s = ",str1,str2);
		for(;i>=0;i--)
		{
		printf("%d",a[i]); // 26行得到的a[i]倒序逐个输出。 
	    }
	}
	else
		printf("0");
	    if(k!=0)
		printf("\n\n");
		if(k==0)
		printf("\n");
	}
}
 
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