一维前缀和——UVA11461 Square Numbers

最新推荐文章于 2022-12-28 11:16:58 发布

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本文介绍了一种通过一维前缀和的方法来高效计算两个指定整数之间包含的所有平方数的数量。利用这种方法可以快速处理大量查询请求，适用于编程竞赛及算法优化场景。

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A - Square Numbers

A square number is an integer number whose square root is also an integer. For example 1, 4, 81 are some square numbers. Given two numbers a and b you will have to find out how many square numbers are there between a and b (inclusive).
Input
The input file contains at most 201 lines of inputs. Each line contains two integers a and b (0 < a ≤ b ≤ 100000). Input is terminated by a line containing two zeroes. This line should not be processed.
Output
For each line of input produce one line of output. This line contains an integer which denotes how many square numbers are there between a and b (inclusive).
Sample Input
1 4
1 10
0 0
Sample Output
2
3

一维前缀和：

a[1],a[2],a[3],......a[n]

s[i]=a[1]+a[2]+a[3]+...a[n],s[0]=0;

a[j]+a[j+1]+....a[k]=s[k]-s[j-1]

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100100;
int s[maxn];
int main()
{
	memset(s,0,sizeof(s));
	for(int i=1,j=1;i<=maxn;i++)
	if(i==j*j)
	{
		s[i]=s[i-1]+1;             //此处前缀和的对象定义为平方数的个数，也是依赖前一项前缀和推来 
		                        //例如定义a数组存放是否为平方数，是为1，则a[10]为1,0,0,1,0,0,0,0,1,0
								//定义s为前缀和数组，s[10]为1,0,0,2,0,0,0,0,3,0      而此处直接使用前缀和即可，不必涉及a数组 （助理解） 
		j++;
	}
	else
	{
		s[i]=s[i-1];
	}
	
	int a,b;
	while(~scanf("%d%d",&a,&b)&&(a||b))//   此处需考虑完整，例如a=0，b=1符合（不应写成a>>0&&b>>0） 
	{
	  printf("%d\n",s[b]-s[a-1]);
	}
	return 0;
}