SPOJ：NT Games（欧拉函数）

Katniss Everdeen after participating in Hunger Games now wants to participate in NT Games (Number Theory Games).

As she begins President Snow provides her a number k. Then, she has to defend t back to back attacks from Haymitch Abernathy for practice. In each attack Haymitch Abernathy gives two numbers l and r, for defense she has to compute :

As she is new to number theory, help her by computing given expression.

Input Format

First line contain an integer, i.e. k.

Second line contain an integer, i.e. t.

Each of next t lines contain two integers, i.e. l & r.

Constraints

1<=k<=10^5

1<=t<=10^5

1<=l<=10^5

l<=r<=10^5

Output Format

For each attack output the value of expression.

Sample Input

1

1

1 5

Sample Output

26

Explanation : Just evaluate the expression.

题意： 求题意的区间的GCD^K之和模Mod

思路：利用前缀和思想+欧拉函数：

Σx  (GCD(i,j)==x，j>i)，枚举X，然后枚举j，根据欧拉函数得到i的数量。

由于询问次数多，我们预处理出答案，预处理的时候，利用前缀和思想降低复杂度。

总的复杂度=N*(N/1+N/2+N/3+N/4+...N/N)=NlogN。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=;
const int Mod=1e9+;
ll phi[maxn+],p[maxn+],vis[maxn+];
ll ans[maxn+],K,T,L,R,cnt;
ll qpow(ll a,ll x){ ll  res=; while(x){ if(x&) res=res*a%Mod; a=a*a%Mod; x>>=;} return res;}
void getphi()
{
    for(ll i=;i<=maxn;i++){
        if(!vis[i]) p[++cnt]=i,phi[i]=i-;
        for(ll j=;j<=cnt&&p[j]*i<=maxn;j++){
            vis[i*p[j]]=;
            phi[i*p[j]]=phi[i]*(p[j]-);
            if(i%p[j]==){
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
        }
    }
}
void solve()
{
    for(ll i=;i<=maxn;i++) ans[i]=(ans[i]+qpow(i,K))%Mod;//自己
    for(ll i=;i<=maxn;i++) ans[i]=(ans[i]+phi[i])%Mod;//
    for(ll i=;i<=maxn;i++){
        for(ll j=;j*i<=maxn;j++)
          ans[i*j]=(ans[i*j]+qpow(i,K)*phi[j]%Mod)%Mod;
    }
    for(ll i=;i<=maxn;i++) ans[i]=(ans[i-]+ans[i])%Mod;
}
int main()
{
    getphi();
    scanf("%lld%lld",&K,&T);
    solve();
    while(T--){
        scanf("%lld%lld",&L,&R);
        printf("%lld\n",((ans[R]-ans[L-])%Mod+Mod)%Mod);
    }
    return ;
}