HDU 1047 Integer Inquiry【大数】

发布时间：2020-12-14 02:27:15 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): 15325????Accepted Submission(s): 3933 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): 15325????Accepted Submission(s): 3933

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

Source

East Central North America 1996

思路：

?????????? 大数多个数相加和一般的数多个数相加方法一样，都是用一个数来保存结果（数的和），将数如的数加到这个结果上直到不满足条件为止，输出结果！区别在于：大数需要用到数组，需要将大数的两个数的加法的算法用到这个上面，具体操作看代码！

代码：

#include <stdio.h>
#include <string.h>
#define N 105
char a[N];
int b[N],c[N];
int main()
{
	int n,i,j,k,len;
	scanf("%d",&n);
	while(n--)
	{
	  memset(c,sizeof(c));//数组c用来保存最终的结果（多个大数的和） 
	  while(gets(a)&&a[0]!='0')
	  {
	  	len=strlen(a);
	  	memset(b,sizeof(b));//数组b用来保存数组a中的字符-‘0’所得到的数值！ 
	  	for(i=len-1,j=0;i>=0;i--)
	  	{
	  		b[j++]=a[i]-'0';
	  	}
	  	for(i=0;i<101;i++)//每次都将b加到c上 
	  	{
	  		c[i]+=b[i];
	  		if(c[i]>=10)
	  		{
	  			c[i]-=10;
	  			c[i+1]++;
	  		}
	  	}
	  }
	  for(i=100;i>=0&&c[i]==0;i--);//输出c！ 
	  if(i>=0)
	    for(;i>=0;i--)
	       printf("%d",c[i]);
	  else
	    printf("0");
	  printf("n");
	  if(n!=0)//两个输出结果之间要空一行！ 
	    printf("n");
	}
	return 0;
}

（编辑：李大同）

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