這倆題目都算是尤拉函式的應用極致了，直接上題目

 Calculation 2

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.InputFor 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.OutputFor each test case, you should print the sum module 1000000007 in a line.Sample Input

3
4
0

Sample Output

0
2

題意：求1-n之間與n不互質的數的和

思路：暴力求解是行不通的，尤拉函式是求與n互質的數的個數，如何求和是個問題

先給出和的公式：sum（n）=n/2*phi[n]

九章算術裡的更相減損法       ↓↓↓↓↓↓↓

簡略的證明一下：gcd（n，k）=gcd（k，n-k）；

使k<n並且gcd（n，k）==1，所以存在gcd（n，n-k）也為1，所以小於n的範圍為之內，出現k和n-k兩組，求所有k的和，即（n/2）*phi[n]

蓋提只需要求出1-n的和來，減去與n互質的數的和即可，最後要mod1000000007

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
#define mod 1000000007
#define ll long long
ll get_phi(ll n)
{
    ll ans=n;
    for(int i=2;i<=(double)sqrt(n*1.0);i++)
    {
        if(n%i==0)
            ans=ans*(1.0-1.0/i);
        while(n%i==0)
            n=n/i;
    }
    if(n>1)
        ans=ans*(1.0-1.0/n);
    return ans;
}
int main()
{
    ll n;
    while(cin>>n)
    {
        if(n==0)
            break;
        ll sum=(n-1)*n/2;
        ll ans=sum-get_phi(n)*n/2;
        cout<<ans%mod<<endl;
    }
    return 0;
}
 

GCD

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, how many integer X satisfies 1<=X<=N and (X,N)>=M.InputThe 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 (2<=N<=1000000000, 1<=M<=N), representing a test case.OutputFor each test case,output the answer on a single line.Sample Input

3
1 1
10 2
10000 72

Sample Output

1
6
260

題意：給n和m，x是1-n之間的數，問你gcd（x，n）>=m的數有多少

思路：剛開始是想從m開始，找到第一個gcd（x，n）>=m的數，然後篩去他的倍數，標記，重複此操作，但是資料太大，不好實現標記

尤拉函式又發揮了

如果m是1，那麼所有的<=n都滿足條件直接出n

m不是1，現x<=n，那假設n=p*d，x=q*d，且gcd（x，n）==d，所以p，q互質，那麼現在問題轉化成了找p使得p*d==n並且d>=m的個數

首先p是n的因子，這個坑的地方就在於n=p*p的時候，留意不要多加

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
#define mod 1000000007
#define ll long long
ll get_phi(ll n)
{
    ll ans=n;
    for(int i=2;i<=(double)sqrt(n*1.0);i++)
    {
        if(n%i==0)
            ans=ans*(1.0-1.0/i);
        while(n%i==0)
            n=n/i;
    }
    if(n>1)
        ans=ans*(1.0-1.0/n);
    return ans;
}
int main()
{
    int t;
    cin>>t;
    ll n,m;
    while(t--)
    {
        cin>>n>>m;
        ll ans=0;
        if(ans==1)
        {
            cout<<get_phi(n)<<endl;
        }
        else
        {
        for(ll i=1;i<=sqrt(n);i++)
        {
            if(n%i==0)
            {
                if(i>=m)
                    ans+=get_phi(n/i);
                if(i*i!=n&&n/i>=m)
                    ans+=get_phi(i);
            }
        }
        cout<<ans<<endl;
    }
    }
    return 0;
}

可以，貼完程式碼的這一瞬見斷了電。。。。。。。。。。。。