hdu 1695 莫比乌斯反演

链接：戳这里

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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).

题意：给出[1,b]和[1,d]两个区间,分别在区间内选x,y使gcd(x,y)==k 求满足的x,y的的对数

思路：当然了  gcd(x/k,y/k)==1  嗯 显然就是莫比乌斯反演了  具体的看上一篇博客把

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 100010
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int a,b,c,d,k;
int mu[MAX],prime[MAX],vis[MAX],cnt;
void Moblus(){
    cnt=0;
    mu[1]=1;
    mst(vis,0);
    for(int i=2;i<=MAX;i++){
        if(!vis[i]){
            mu[i]=-1;
            prime[++cnt]=i;
        }
        for(int j=1;j<=cnt;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(){
    Moblus();
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        printf("Case %d: ",cas);
        if(k==0){
            cout<<0<<endl;
            continue;
        }
        if(b>d) swap(b,d);
        b/=k;d/=k;
        ll ans1=0,ans2=0;
        for(int i=1;i<=b;i++){
            ans1+=(ll)mu[i]*(b/i)*(d/i);
        }
        for(int i=1;i<=b;i++){
            ans2+=(ll)mu[i]*(b/i)*(b/i);
        }
        printf("%I64d\n",ans1-(ans2/2));
    }
    return 0;
}