题目见 http://pan.baidu.com/s/1o6zajc2
此外不知道H-L<=10^5这个条件是干嘛的。。。。
#include <map> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define M 10001000 #define INF 0x3f3f3f3f #define MOD 1000000007 using namespace std; int mu[M],prime[1001001],tot; bool not_prime[M]; map<int,long long> mu_sum; long long n,d,l,r; void Linear_Shaker() { int i,j; mu[1]=1; for(i=2;i<=10000000;i++) { if(!not_prime[i]) { mu[i]=-1; prime[++tot]=i; } for(j=1;prime[j]*i<=10000000;j++) { not_prime[prime[j]*i]=true; if(i%prime[j]==0) { mu[prime[j]*i]=0; break; } mu[prime[j]*i]=-mu[i]; } } for(i=1;i<=10000000;i++) mu[i]+=mu[i-1]; } long long Mu_Sum(int x) { if(x<=10000000) return mu[x]; if(mu_sum.find(x)!=mu_sum.end()) return mu_sum[x]; long long i,last,re=1; for(i=1;i<=x;i=last+1) { last=x/(x/i); if(x/i-1) re-=(Mu_Sum(last)-Mu_Sum(i-1))*(x/i-1); } return mu_sum[x]=re; } long long Quick_Power(long long x,int y) { long long re=1; while(y) { if(y&1) (re*=x)%=MOD; (x*=x)%=MOD; y>>=1; } return re; } long long Solve() { long long i,last,re=0; for(i=1;i<=r;i=last+1) { last=min(r/(r/i),l/i?(l/(l/i)):INF); re+=(Mu_Sum(last)-Mu_Sum(i-1))*Quick_Power(r/i-l/i,n); re%=MOD; } return (re%MOD+MOD)%MOD; } int main() { cin>>n>>d>>l>>r; l=(l-1)/d;r=r/d; Linear_Shaker(); cout<<Solve()<<endl; }
时间: 2024-10-29 19:06:58