HDU 1695：GCD _容斥原理

最新推荐文章于 2021-01-31 16:30:00 发布

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本文探讨了在给定区间内寻找满足特定条件的整数对的方法，具体为找到所有使得最大公约数等于给定值的整数对。通过转换问题，将原始问题转化为求解区间内互质数对的数量，并应用欧拉函数和容斥原理求解。

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GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2474 Accepted Submission(s): 902

Problem Description

Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

Output

For each test case, print the number of choices. Use the format in the example.

Sample Input

  
   2
1 3 1 5 1
1 11014 1 14409 9

Sample Output

  
   Case 1: 9
Case 2: 736427


   
    
     Hint
    For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

Source

2008 “Sunline Cup” National Invitational Contest

Recommend

wangye

//本题没想到好的方法，借用了某牛的思路才写过的！

//分析：题目是要求a<=x<=b,c<=y<=d(其中a,c题目中已经说了必为1)，中有多少对(x,y)（无序的）满足GCD（x,y）=k，那么可以转化成求（x,y）分别在区间1<=x<=b/k,1<=y<=d/k中有多少对满足GCD(x,y)=1；转换b/=k,d/=k，之后，假设b<=d，那么我们可以分两步来求值：

1、求[1,b]与[1,b]之间有多少对数互质，显然就是phi[1]+phi[2]+...phi[b]即可，用线性求欧拉函数的算法即可！

2\求[1,b]与[b+1,d]之间有多少对数互质，可以先求出有多少对数不互质，然后减去即可！针对于区间[b+1,d]中得某一个数n求区间[1,b]中有多少个数与他不互质（即有公约数）那么可以用容斥原理较好的解决！

两步求值之和即为题意所求！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
#define maxn 100005

int phi[maxn],x[maxn],cnt[1005];
void Init()
{
	int i,j;
	memset(x,0,sizeof(x));
	x[0]=x[1]=1;
	for(i=1;i<maxn;i++)
		phi[i]=i;
	for(i=2;i<maxn;i++)
	{
		if(!x[i])
		{
			phi[i]=i-1;
			for(j=i*2;j<maxn;j+=i)
			{
				x[j]=1;
				phi[j]=phi[j]/i*(i-1);
			}
		}	
	}
}

LL Imco_prime(int n,int m)
{
	int i,j,t=0;
	for(i=2;i*i<=n;i++)
	{
		if(n&&n%i==0)
		{
			cnt[t++]=i;
			while(n&&n%i==0)
				n/=i;
		}
	}
	if(n>1)
		cnt[t++]=n;
	LL ans=0,tmp,flag;
	for(i=1;i<(1<<t);i++)
	{
		tmp=1,flag=0;
		for(j=0;j<t;j++)
			if(i&(1<<j))
				flag++,tmp*=cnt[j];
		if(flag&1)
			ans+=m/tmp;
		else
			ans-=m/tmp;
	}
	return ans;
}

int main()
{
	Init();
	int T,t=0,a,b,c,d,k,i,j;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
		if(k==0)
		{
			printf("Case %d: 0\n",++t);
			continue;
		}
		b/=k,d/=k;
		if(b>d)
			swap(b,d);
		LL ans=0,tmp=(LL)b*(d-b);  //细节
		for(i=1;i<=b;i++)
			ans+=phi[i];
		for(i=b+1;i<=d;i++)
			tmp-=Imco_prime(i,b);
		printf("Case %d: %I64d\n",++t,ans+tmp);	
	}
	return 0;
}