标签：not int ane str lld which 欧拉 test 函数
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
如果gcd(n,i)==1,gcd(n,n-i)==1
证：i=1(mod n)
-i=-1(mod n)
n-i=-1+n (mod n)
n-i=1 (mod n)
通过欧拉函数一个数n中存在phi(n)个与之互质的数，因为i+（n-i）为n也就是有n对。
则与n互质数目之和为res=phi(n)/2*n,
ans=(n-1)*n/2-res。
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define N 1000010
#define maxn 1000010
#define mod 1000000007
using namespace std;
typedef long long ll;
int p[N];
int prime[N];
int pn=0;
bool vis[N];
int main()
{
for (int i = 2; i < N; i++) {
if (vis[i]) continue;
prime[pn++] = i;
for (int j = i; j < N; j += i)
vis[j] = 1;
}
ll n;
while(~scanf("%lld",&n),n)
{
ll r=n;
ll phi=n;
for(int i=0;prime[i]*prime[i]<=r;i++)
{
ll tem=0;
while(r%prime[i]==0)
{
r/=prime[i];
tem++;
}
if(tem)
phi=phi-phi/prime[i];
}
if(r!=1)
phi-=phi/r;
ll rem=n*phi/2;
ll ans=n*(n-1)/2;
printf("%lld\n",(ans-rem)%mod);
}
}
标签：not int ane str lld which 欧拉 test 函数
原文地址：http://www.cnblogs.com/ygtzds/p/7840569.html
