Problem Description
The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+….+ (b)
Input
There are several test cases. Each line has two integers a, b (2<a<b<3000000).
Output
Output the result of (a)+ (a+1)+….+ (b)
数学-线性欧拉函数+前缀和
前缀和用long long
点击查看代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=3e6+10;
int phi[N],p[N/5];
ll s[N];
bool vis[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int a,b;
phi[1]=1;s[1]=1;
int cnt=0;
for(int i=2;i<N;++i)
{
if(!vis[i]) p[++cnt]=i,phi[i]=i-1;
for(int j=1;j<=cnt&&i*p[j]<N;++j)
{
vis[i*p[j]]=1;
if(i%p[j]==0) {phi[i*p[j]]=p[j]*phi[i];break;}
phi[i*p[j]]=(p[j]-1)*phi[i];
}
s[i]=s[i-1]+phi[i];
}
while(cin>>a>>b)
{
cout<<s[b]-s[a-1]<<'\n';
}
return 0;
}