hdu GCD(莫比乌斯反演+一点学习笔记)_莫比乌斯反演.pptx

本文链接：https://blog.youkuaiyun.com/sudu6666/article/details/82427160

GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15999 Accepted Submission(s): 6152

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

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题意: $\sum_{x=1}^{b}\sum_{y=1}^{d}[gcd(x,y)==k]$ 的x,y的对数

题解:还是第一次涉及到莫比乌斯反演的题目,这里就先写一下对于莫比乌斯反演的一些理解吧,首先莫比乌斯反演是一种化简公式从而简化运算的一种方法,比如问题让我们求解 F(n) ,那么我们可以先定义一个函数 G(n) ,使得 $G(n)=\sum_{d|n}F(d)$ ,因为G(n)我们已知,那么就可以通过莫比乌斯反演公式 $F(n)=\sum_{d|n}\mu(n/d)G(d)$ 求出 F(n) 了。( $\mu$ 是一个容斥因子)

$\sum_{x=1}^{b}\sum_{y=1}^{d}[gcd(x,y)==k]$ $G(n) = \sum_{x=1}^{b/k}\sum_{y=1}^{d/k}[n|gcd(x,y)]$

$=\sum_{x=1}^{b}\sum_{y=1}^{d}[gcd(x/k,y/k)=1]$ $=F(n)=\sum_{n|d}\mu(d/n)*G(d)$

$=\sum_{x=1}^{b/k}\sum_{y=1}^{d/k}[gcd(x,y)=1]$ 即为答案

#include<stdio.h>
#include<algorithm>
#include<iostream>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define SI(i) scanf("%lld",&i)
#define PI(i) printf("%lld\n",i)
using namespace std;
typedef long long ll;
const int mod=998244353;
const int MAX=1e6+7;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
int dir[9][2]={0,1,0,-1,1,0,-1,0, -1,-1,-1,1,1,-1,1,1};
template<class T>bool gmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>bool gmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>void gmod(T &a,T b){a=((a+b)%mod+mod)%mod;}
typedef pair<ll,ll> PII;

int mu[MAX],vis[MAX],prime[MAX];
void get_mobi()
{
    mu[1]=1;
    int tot=0;
    for(int i=2;i<MAX;i++)
    {
        if(!vis[i])
        {
            prime[++tot]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=tot&&i*prime[j]<MAX;j++)
        {
            if(i*prime[j]>MAX) break;
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                 mu[i*prime[j]]=0;
                 break;
            }
            else mu[i*prime[j]]=-mu[i];
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    get_mobi();
    for(int cas=1;cas<=T;cas++)
    {
        int a,b,c,d,k;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);

        printf("Case %d: ",cas);

        if(!k)
        {
            printf("0\n");
            continue;
        }

        int N=min(b,d);
        b/=k;d/=k;
        if(b>d) swap(b,d);

        ll ans=0,tmp=0;
        for(int i=1;i<=N;i++)
        {
            ans+=1LL*mu[i]*(b/i)*(d/i);
            tmp+=1LL*mu[i]*(b/i)*(b/i);
        }

        printf("%lld\n",ans-tmp/2);
    }

    return 0;
}