HDU 3501 Calculation 2 (尤拉函式應用)

_TCgogogo_發表於2015-08-25
函式


Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2989    Accepted Submission(s): 1234


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




 

Author 

GTmac


 

Source 

2010
ACM-ICPC Multi-University Training Contest（7）——Host by HIT 



 


題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=3501


題目大意：求小於n的與n不互質的數的和


題目分析：考慮到小於n的與n互質的數的和很好求，再用總和減掉即可，小於n的與n互質的數的和等於n*phi[n] / 2，phi[n]為n對應的尤拉函式值，證明如下：

設gcd(n, i) == 1，則有gcd(n, n - i) == 1

這裡可以反證：假設存在一個k != 1，gcd(n, n - i) == k，則有n%k == 0且(n - i) % k == 0即(n % k - i % k) % k == 0，得到(i % k) % k == 0

因此i必然是k的倍數，所以gcd(n, i) == k，這與gcd(n, i) == 1衝突，因此對於gcd(n, i) == 1，有gcd(n, n - i) == 1，所以與n互質的兩對數相加為n，又與n互質的總個數為phi[n]，注意這裡不會出現n == 2*i 的情況因為出了2，phi[i]都為偶數

所以小於n的與n互質的數的和等於n*phi[n] / 2，所以最後答案為(n * (n - 1) / 2 - n*phi[n] / 2) % MOD，求單個phi的複雜度為sqrt((n)

#include <cstdio>
#include <cmath>
#define ll long long
int const MOD = 1e9 + 7;

int phi(int x)
{
    int res = x;
    for(int i = 2; i * i <= x; i++)
    {
        if(x % i == 0) 
        {
            res = res / i * (i - 1);
            while(x % i == 0) 
                x /= i;
        }
    }
    if(x > 1) 
        res = res / x * (x - 1);
    return res;
}

int main()
{
    int n;
    while(scanf("%d", &n) != EOF && n)
    {
        ll sum = (ll) n * (n - 1) / 2;
        ll copsum = (ll) n * phi(n) / 2;
        ll ans = sum - copsum;
        printf("%I64d\n", ans % MOD);
    }
}












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