题解:

莫比乌斯函数

然而向我这种弱菜肯定选择暴力dp

代码:

#include<bits/stdc++.h>
const int N=100010,M=1000000007;
typedef long long ll;  
using namespace std;  
int n;  
ll m,L,R,k,f[N];   
ll power(ll x,int y)  
{  
    ll ans=1;  
    while (y)  
     {  
        if (y&1)ans=ans*x%M;  
        x=x*x%M;  
        y>>=1;  
     }  
    return ans;  
}  
int main()  
{  
    scanf("%d%lld%lld%lld",&n,&k,&L,&R);  
    for (ll i=R-L;i>=1;i--)  
     {  
        ll l=(L-1)/(k*i),r=R/(k*i);  
        f[i]=(power(r-l,n)-(r-l)+M)%M;  
        for (int j=2;i*j<=R-L;j++)f[i]=(f[i]-f[i*j]+M)%M;  
     }  
    if (L<=k&&k<=R)f[1]++; 
    printf("%d\n",f[1]);  
    return 0;  
}  

 

posted on 2018-03-28 17:29  宣毅鸣  阅读(79)  评论(0编辑  收藏  举报