GCD nyoj1007（欧拉函数运用&&数论入门）

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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

31 110 210000 72

样例输出

1351305000

上传者

ACM_张书军

题意：Given integers N and M,please answer sum of X satisfies 1<=X<=N and (X,N)>=M./*就在这一句了*/

给你两个数 N M 求1~N 之间所有gcdx的和

/*在数论，对正整数n，欧拉函数是少于或等于n的数中与n互质的数的数目。*//*思路：枚举n的因子。假设n的因子为d。d*gcd(x/d,n/d)=1。d*Euler(n/d)就是因子为gcd(x,n)=d，从而求gcd(x,n)的和。*/#include<stdio.h>#include<string.h>#include<iostream>using namespace std;const int mod=1000000007;long long Euler(long long n)//欧拉函数{    long long c=n,i;    for(i=2; i*i<=n; i++)    {        if(n%i==0)        {            while(n%i==0) n/=i;            c=c/i*(i-1);//φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pn);        }    }    if(n!=1)        c=c/n*(n-1);    return c;}//求 x和long long Euler_sum(long long n){    if(n==1)        return 1;    else        return n*Euler(n)/2;}int main(){    long long  a,b;    int t;    cin>>t;    while(t--)    {        while(cin>>a>>b)        {            int cnt;            long long i,c=0;            for(i=1; i*i<=a; i++)            {                if(a%i==0)                {                    if(i>=b)                    {                        cnt=i;//- -                        c=(c+cnt*Euler_sum(a/cnt))%mod;                    }                    if(i*i!=a&&a/i>=b)//枚举i与n的因子。                    {                        cnt=a/i;                        c=(c+cnt*Euler_sum(a/cnt))%mod;                    }                }            }            cout<<c<<endl;        }    }}

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