FZU 1969 && UVA 11426 GCD Extreme (尤拉函式 或 莫比烏斯反演)

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GCREM函式
GCD Extreme

Accept: 124    Submit: 249
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

Given the value of N, you will have to find the value of G. The meaning of G is given in the following code 

G=0; 
for(i=1;i<N;i++)
    for(j=i+1;j<=N;j++) 

        G+=gcd(i,j);

/*Here gcd() is a function that finds the greatest common divisor of the two input numbers*/

Input

The input file contains at most 20000 lines of inputs. Each line contains an integer N (1<N <1000001). The meaning of N is given in the problem statement. Input is terminated by a line containing a single zero.

Output

For each line of input produce one line of output. This line contains the value of G for the corresponding N. The value of G will fit in a 64-bit signed integer.

Sample Input

10
100
200000
0

Sample Output

67
13015
143295493160

Source

Contest for 2010 lecture II  

題目連結：http://acm.fzu.edu.cn/problem.php?pid=1969

題目大意：求題面程式段的值

題目分析：先看兩個資料和時限，在Uva上本題100組資料，n為4e6，時限10s，Fzu上20000組資料，n為1e6，時限1s，先說複雜度高的莫比烏斯反演的做法，對於每個輸入的n，列舉最大公約數然後利用反演容斥，複雜度O(Case * nlogn)，在Uva上3s＋跑完
#include <cstdio>
#define ll long long
int const MAX = 4e6 + 5;
int mob[MAX], p[MAX], sum[MAX];
bool noprime[MAX];

void Mobius()
{
    int pnum = 0;
    mob[1] = 1;
    for(int i = 2; i < MAX; i++)
    {
        if(!noprime[i])
        {
            p[pnum ++] = i;
            mob[i] = -1;
        }
        for(int j = 0; j < pnum && i * p[j] < MAX; j++)
        {
            noprime[i * p[j]] = true;
            if(i % p[j] == 0)
            {
                mob[i * p[j]] = 0;
                break;
            }
            mob[i * p[j]] = -mob[i];
        }
    }
}

int main()
{
    Mobius();
    int n;
    while(scanf("%d", &n) != EOF && n)
    {
        ll ans = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = i; j <= n; j += i)
                ans += (ll)i * (n / j) * (n / j) * mob[j / i];
            ans -= (ll) i;
        }
        printf("%lld\n", ans / 2);
    }
}


FZU上這種做法肯定超時，回到題目，因為有∑gcd(i, N) = ∑(d|N) d*phi[N/d]，則G(N) = ∑gcd(1, N) + ∑gcd(2, N) + ... + ∑gcd(N-1, N)
令f[i] = ∑gcd(i, N)，則G(N) = ∑f[i]，只需要篩出f[i]，求字首和預處理然後就能O(1)查詢，複雜度為O(MAX*logMAX)，100ms＋，FZU上的long long要用I64d
#include <cstdio>
#define ll long long
int const MAX = 1000005;
int p[MAX], phi[MAX], f[MAX], n;
ll ans[MAX];
bool noprime[MAX];

void pre()  
{  
    int pnum = 0;
    for(int i = 2; i < MAX; i++)  
    {  
        if(!noprime[i])  
        {  
            p[pnum ++] = i;  
            phi[i] = i - 1;  
        }  
        for(int j = 0; j < pnum && i * p[j] < MAX; j++)  
        {  
            noprime[i * p[j]] = true;
            if(i % p[j] == 0)  
            {  
                phi[i * p[j]] = phi[i] * p[j];  
                break;  
            }  
            phi[i * p[j]] = phi[i] * (p[j] - 1);  
        }  
    }  
    for(int i = 1; i < MAX; i++)
    {
        for(int j = i; j < MAX; j += i)
            f[j] += (ll)i * phi[j / i];
        ans[i] = ans[i - 1] + f[i];
    }
}  

int main()
{
    pre();
    while(scanf("%d", &n) != EOF && n)
        printf("%I64d\n", ans[n]); 
}



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