uva 11317 - GCD+LCM(欧拉函数+log)

题目链接：uva 11317 - GCD+LCM

题目大意：给定n，求出1~n里面两两的最大公约的积GCD和最小公倍数的积LCM，在10100进制下的位数。

解题思路：在n的情况下，对于最大公约数为i的情况又phi[n/i]次。求LCM就用两两乘积除以GCD即可。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1000000;
const double logt = log(10.0);

int N;
double g[maxn+5], s[maxn+5];
int phi[maxn+5];

void phi_table (int n) {
    memset(phi, 0, sizeof(phi));
    phi[1] = 1;

    for (int i = 2; i <= n; i++) {
        if (!phi[i]) {
            for (int j = i; j <= n; j += i) {
                if (!phi[j])
                    phi[j] = j;
                phi[j] = phi[j] / i * (i-1);
            }
        }
    }
}

void init (ll n) {

    for (int i = 1; i <= n; i++) {
        double tmp = log(i);
        for (int j = 2 * i; j <= n; j += i)
            g[j] += phi[j/i] * tmp;
    }

    for (int i = 1; i <= n; i++)
        g[i] += g[i-1];
    for (int i = 1; i <= n; i++)
        g[i] /= logt;
}

int main () {
    phi_table(maxn);
    init(maxn);

    int cas = 1;
    while (scanf("%d", &N) == 1 && N) {
        double ans = 0;
        for (int i = 1; i <= N; i++)
            ans += (N-1) * log(i);
        ans /= logt;
        ans -= g[N];
        printf("Case %d: %lld %lld\n", cas++, (ll)(g[N] / 100) + 1, (ll)(ans / 100) + 1);
    }
    return 0;
}