G

原题链接

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)

Sample Input
3 100

Sample Output
3042
题意分析：求a到b之间所有的欧拉函数值的和
相当于一个模版题，先筛出1-maxn的所有欧拉函数的值即可，文末有线性的欧拉筛，更快

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
#include <math.h>
#include <map>
#include <queue>
#include <set>
using namespace std;
typedef long long ll;
const int maxn=3e6+7;
const int mod=1e9+7;
ll p[maxn];
//o(nloglogn)的筛法
void get_Phi()
{
    for(int i=2;i<=maxn;i++)p[i]=0;
        p[1]=1;
    for(int i=2;i<=maxn;i++)
        if(!p[i]){
            for(int j=i;j<=maxn;j+=i){
                if(!p[j])p[j]=j;
            p[j]=p[j]/i*(i-1);
           }
        }
}
int main()
{
    //freopen("in.txt","r",stdin);
    get_Phi();
    int a,b;
    while(scanf("%d%d",&a,&b)!=-1)
    {
        ll ans=0;
        for(int i=a;i<=b;i++)
            ans+=p[i];
        printf("%lld
",ans );
    }
    //printf("%lld
",ans );
    return 0;
}

//线性筛
const int N=3e6+7;
int prime[N],isprime[N];//记录素数的
int phi[N];
void get_phi(){
    int i,j,cnt=0;
    for (i=2; i<=N; i++) {
        if (isprime[i]==0){
            prime[cnt++]=i;
            phi[i]=i-1;
        }
        for (j=0; j<cnt&&i*prime[j]<N; j++) {
            isprime[i*prime[j]]=1;
            if (i%prime[j]==0) {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }else
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
}