P1348 Couple number

P1348 Couple number

解法一：

粉色骚一点。

求完全平方数的个数，需要注意的是：有负数。

C=(a+b)(a-b).

a+b和a-b显然具有相同的奇偶性

C=2k*2i=4*k*i=4的倍数

C=(2k+1)*(2i+1)=4*k*i+2k+2i+1=偶数+奇数(2i+1)=奇数

证明：

N=4K=2K*2=(k+1+k-1)*(k+1-(k-1))=(2k)*2=4k=>a=k+1 b=k-1

N=2k+1=(2k+1)*1=(k+1+k)*(k+1-k)=2k+1=>a=k+1 b=k

#include<string>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<map>
#include<vector>
#include<set>

#define FOR(a,b) for(register int i=a;i<=b;i++)
#define ULL  unsigned long long
#define  LL  long long

using namespace std;


int main()
{
	LL n1, n2;
	cin >> n1 >> n2;
	int ans = 0;
	for (LL i = n1; i <= n2; i++)
	{
		if (i % 4 == 0 || i % 2 != 0)
		{
			ans++;
		}
	}
	cout << ans << endl;
}

解法二：

N=(A+B)*(A-B);

N=4n=(k+1+k-1)*(2)=4k =>a=k+1 b=k-1

N=4n+1=(2n+1+2n)*(1) =>a=2n+1 b=2n

N=4n+2=2(2n+1)=奇数*偶数 显然不符合要求

N=4n+3=(2n+2+2n+1)*(2n+2-2n-1) =>a=2n+2 b=2n-1

筛去4n+2的，剩下的都满足条件

#include<string>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<map>
#include<vector>
#include<set>

#define FOR(a,b) for(register int i=a;i<=b;i++)
#define ULL  unsigned long long
#define  LL  long long

using namespace std;


int main()
{
	LL n1, n2;
	cin >> n1 >> n2;
	int ans = 0;
	for (LL i = n1; i <= n2; i++)
	{
		if (abs(i) % 4 != 2)
		{
			ans++;
		}
	}
	cout << ans << endl;
}

参考：

https://www.luogu.org/problemnew/solution/P1348