hdu 2824 The Euler function 欧拉函数打表

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

思路：欧拉函数打表；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
int p[3000010],N=3000010;
void phi()
{
    for(int i=1; i<N; i++)  p[i] = i;
    for(int i=2; i<N; i+=2) p[i] >>= 1;
    for(int i=3; i<N; i+=2)
    {
        if(p[i] == i)
        {
            for(int j=i; j<N; j+=i)
            p[j] = p[j] - p[j] / i;
        }
    }
}
int main()
{
    int x,y,z,i,t;
    phi();
    while(~scanf("%d%d",&x,&y))
    {
        ll ans=0;
        for(i=x;i<=y;i++)
        ans+=p[i];
        printf("%I64d
",ans);
    }
    return 0;
}