1.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
题意是找出小于n的与n互质数之和。互质也就是两者最大公约数为1
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
ll phi(ll n)
{
ll res=n;
for(ll i=2;i*i<=n;i++)
{
if(n%i==0)
{
res=res/i*(i-1);
while(n%i==0)
n/=i;
}
}
if(n>1)res=res/n*(n-1);
return res;
}
int main()
{
ll n;
while(cin>>n&&n)
{
ll m=phi(n);
ll k=m*n/2;//k是满足条件与n互素数的和
k=n*(n+1)/2-n-k;//n*(n+1)/2为1---n所有数的和
cout<<k%1000000007<<endl;
}
}
2.给你一个整数N，求范围小于N中的整数中，与N的最大公约数大于1的整数的个数。
欧拉函数求出的是小于n中与n最大公约数为1的个数所以
ans = n - φ(N) - 1。
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
ll phi(ll n)
{
ll res=n;
for(ll i=2;i*i<=n;i++)
{
if(n%i==0)
{
res=res/i*(i-1);
while(n%i==0)
n/=i;
}
}
if(n>1)res=res/n*(n-1);
return res;
}
int main()
{
ll n;
while(cin>>n&&n)
{
ll m=phi(n);
ll k=n-m-1;
cout<<k<<endl;
}
}
也可以用筛法做
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
const ll N =100000000;
ll phi[N+10],prime[N+10],tot,ans;//用筛法要注意数组不要越界，注意n的范围
bool mark[N+10];
void ok()
{
int i,j;
phi[1]=1;
for(i=2;i<=N;i++)
{
if(!mark[i])
{
prime[++tot]=i;
phi[i]=i-1;
}
for(j=1;j<=tot&&i*prime[j]<=N;j++)
{
mark[i*prime[j]]=1;
if(i%prime[j]==0)
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
else
{
phi[i*prime[j]]=phi[i]*(prime[j]-1);
}
}
}
}
int main()
{
ok();
ll n;
while(cin>>n)
{
ll m=phi[n];
ll k=n-m-1;
cout<<k<<endl;
}
}