nyoj1007——欧拉求和

GCD

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

 

描述

The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M,please answer sum of  X satisfies 1<=X<=N and (X,N)>=M.

 

输入

The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (1<=N<=10^9, 1<=M<=10^9), representing a test case.

输出

Output the answer mod 1000000007

样例输入

3
1 1
10 2
10000 72

样例输出

1
35
1305000

上传者

ACM_张书军

#include <bits/stdc++.h>
using namespace std;
typedef long long  ll;
const int INF = 0x3f3f3f3f;
const int maxn = 2000;
const int moder = 1000000007;

ll eular(ll n)
{
    ll ans = n;
    for(int i=2;i*i <= n;i++){
        if(n%i == 0){
            ans = ans - ans/i;
            while(n%i == 0)
                n = n/i;
        }
    }
    if(n > 1)
        ans = ans - ans/n;
    return ans;
}

ll eular_sum(ll k)
{
    if(k==1||k==2)
        return 1;
    else return k*eular(k)/2;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        ll n,m;
        scanf("%lld%lld",&n,&m);
        ll res = 0;
        for(int i=1;i*i <= n;i++) {
            if (n % i == 0) {
                if (i >= m) {
                    res = (res + i * eular_sum(n / i))%moder;
                }
                if (i * i != n && n / i >= m){
                    res = (res + n/i*eular_sum(i))%moder;
                }
            }
        }
        printf("%lld
",res);
    }
    return 0;
}

//求欧拉函数（即n以内所有与n互质的数的个数） 

// 设n的质因数分别为p1，p2,.....，pn 

//f(x)=n*(1-p1)*(1-p2)*...*(1-pn)

//求与n互质的数之和S(x)=f(x)/2*x