HDU

题目：求出小于N的与N不互质的数的和。

思路：小于N的与N互质的数的和为eular(n)*n/2

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fconst LL MOD=1000000007;LL phi(LL n){    LL res=n;    for(LL i=2;i*i<=n;i++)        if(n%i==0){            res=res-res/i;            while(n%i==0){                n/=i;            }        }    if(n>1)        res=res-res/n;    return res;}int main(){    LL n;    while(~scanf("%lld",&n)){        if(!n)            break;        LL ans=(n*(n+1)/2-n)%MOD;        ans-=(phi(n)*n)/2%MOD;        ans=(ans+MOD)%MOD;        printf("%lld\n",ans);    }    return 0;}