HDOJ 1047 Integer Inquiry(多次大数相加)

发布时间：2020-12-14 02:27:50 所属栏目：大数据来源：网络整理

导读：Integer Inquiry Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15140????Accepted Submission(s): 3889 Problem Description One of the first users of BIT's new supercomputer was Chip D

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15140????Accepted Submission(s): 3889

Problem Description

One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment,once one became available on the third floor of the Lemon Sky apartments on Third Street.)

Input

The input will consist of at most 100 lines of text,each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length,and will only contain digits (no VeryLongInteger will be negative).

The final input line will contain a single zero on a line by itself.

Output

Your program should output the sum of the VeryLongIntegers given in the input.

This problem contains multiple test cases!

The first line of a multiple input is an integer N,then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

  
  
   
   1 123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0

Sample Output

  
  
   
   370370367037037036703703703670

不知道是题目描述太模糊，还是我英语太差，半天没看懂。题意是：输出一个n，表示有n组测试数据，每组测试数据由多个长度不大于100的大数组成，每组测试数据由0结束，求这些大数相加之和。每组输出结果与下组输入实例之间空一行。

典型的大数题，很简单，大数A+B的变形，注意格式，及00+00的情况就好

代码如下：

#include<cstdio>
#include<cstring>
int a[110];
char s[110],str[110];
void add(char s[],int a[])
{
	int b[110],i,len,j;
	memset(b,sizeof(b));
	len=strlen(s);
	for(i=0;i<len;i++)
	   b[len-i-1]=s[i]-'0';
	for(i=0;i<len;i++)
	{
		a[i]+=b[i];
		if(a[i]>9)
		{
			a[i+1]++;
			a[i]-=10;
		}
	}
}

int main()
{
	int n,j;
	scanf("%d",&n);
	while(n--)
	{
		memset(a,sizeof(a));
		while(scanf("%s",&str)&&strcmp(str,"0")!=0)
			add(str,a);
		for(i=110;i>0;i--)//此处i!=0,防止出现结果为0，要输出0的情况 
		  if(a[i])
		    break;
		for(;i>=0;i--)
		   printf("%d",a[i]);
		printf("n");
		if(n)
		  printf("n");
	}
	return 0;
}

（编辑：李大同）

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