HDU 1695 GCD （数论-整数和素数，组合数学-容斥原理）

GCD

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

题目大意：

从1~a区间取一个数x,从1~b区间取一个数y，问你gcd(x,y)=k有多少种方案？其中x1,y1和y1,x1算同一种方案。

解题思路：

那么就是从1~b/k 取一个数x ，与从1~d/k 取一个数y 互质的方案数，利用容斥，枚举 x，求出y的个数即可。

解题代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;

typedef long long ll;
const int maxn=450;
bool isPrime[maxn];
vector <int> v;
int tol,a,b,c,d,k;

void get_prime(){
    memset(isPrime,true,sizeof(isPrime));
    for(int i=2;i<maxn;i++){
        if(isPrime[i]){
            tol++;
            v.push_back(i);
        }
        for(int j=0;j<tol && i*v[j]<maxn;j++){
            isPrime[i*v[j]]=false;
            if(i%v[j]==0) break;
        }
    }
}

inline vector <int> getPrime(ll x){
    vector <int> tmp;
    for(ll i=0;i<tol && x>=v[i]*v[i];i++){
        if(x%v[i]==0){
            tmp.push_back(v[i]);
            while(x>0 && x%v[i]==0) x/=v[i];
        }
    }
    if(x>1) tmp.push_back(x);
    return tmp;
}

inline ll getCnt(int i,int y){//1-y与i不互质的数的个数
    vector <int> tmp=getPrime(i);
    int vsize=tmp.size();
    ll sum=0;
    for(int t=1;t<(1<<vsize);t++){
        int cnt=0,all=1;
        for(int j=0;j<vsize;j++){
            if(t&(1<<j)){
                all*=tmp[j];
                cnt++;
            }
        }
        if(cnt&1) sum+=y/all;
        else sum-=y/all;
    }
    return sum;
}

void solve(){
    if(k==0 || k>b || k>d){
        cout<<0<<endl;
        return;
    }
    int x=b/k,y=d/k;
    if(x<y) swap(x,y);
    ll ans=0;
    for(int i=1;i<=x;i++){
        int r=i>y?y:i;
        ans+=r;
        ans-=getCnt(i,r);
    }
    cout<<ans<<endl;
}

int main(){
    get_prime();
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        printf("Case %d: ",i);
        solve();
    }
    return 0;
}

HDU 1695 GCD （数论-整数跟素数，组合数学-容斥原理）

GCD

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