【HDU 3501 Calculation 2】+ 欧拉函数

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本文介绍了一种计算小于N且与N不互质的所有正整数之和的方法，并提供了一个基于欧拉函数改进的算法实现。通过该算法，可以高效地解决这一数学问题。

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Calculation 2
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3729 Accepted Submission(s): 1564

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：求1 ~ N 中与 N 不互质的数的总和
欧拉公式的引伸：小于或等于n的数中，与n互质的数的总和为：φ(x) * x / 2。(n>=1)；

#include<cstdio>
typedef long long LL;
const LL mod = 1e9 + 7;
LL oula(LL x){
    LL cut = x;
    for(LL i = 2 ; i * i <= x; i++){
        if(x % i == 0){
            cut = cut / i * (i - 1);
            while(x % i == 0)
                x /= i;
        }
    }
    if(x > 1)
        cut = cut / x * (x - 1);
        return cut;
}
int main()
{
    LL N;
    while(scanf("%lld",&N)!=EOF && N){
    LL ans = (N * (N - 1) / 2 - oula(N) * N / 2) % mod;
    printf("%lld\n",ans);
    }
    return 0;
}