hdu 3501 Calculation 2 欧拉函数

Calculation 2

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

Problem Description

Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.

 

Input

For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.

 

Output

For each test case, you should print the sum module 1000000007 in a line.

 

Sample Input

3 4 0

 

Sample Output

0 2

 

Author

GTmac

 

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

思路：没接触欧拉函数的思路就是，利用唯一分解定理；得到所有质因数；

　　   然后再利用容斥原理得到答案；

　　   然后有个欧拉函数；

　　　首先你需要知道有个这样的定理:如果 gcd(n,i)=1则 gcd(n,n-i)=1 (1<=i<=n)

　　   可得小于n的并与n互质的和为p*phi(p)/2;

　　　这题需要求总和再相减；

#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 ;
}
ll phi(ll n)
{
    ll i,rea=n;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            rea=rea-rea/i;
            while(n%i==0)  n/=i;
        }
    }
    if(n>1)
        rea=rea-rea/n;
    return rea;
}
int main()
{
    ll x;
    while(~scanf("%I64d",&x))
    {
        if(!x)break;
        ll ans=(x*(x-1-phi(x))/2);
        printf("%I64d
",ans%mod);
    }
    return 0;
}