BZOJ 2190 仪仗队（欧拉函数）

题目连接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=47144

题意：求gcd(x, y) = 1,  0 <= x, y <= n的对数

思路：这题范围是从0开始的，所以，会多出(1, 0)和(0, 1)这两组，答案就是sigma(phi(i)) - 1 + 2。

code：

#include <cstdio>
#include <cstring>
typedef long long LL;
const int MAXN = 100005;

LL phi[MAXN];
int primes[MAXN];
bool is[MAXN];
int cnt;


void init()
{
    phi[1] = 1L;
    cnt = 0;
    memset(is, false, sizeof(is));
    for (int i = 2; i < MAXN; ++i) {
        if (!is[i]) {
            primes[cnt++] = i;
            phi[i] = i - 1;
        }
        for (int j = 0; j < cnt && i * primes[j] < MAXN; ++j) {
            is[i * primes[j]] = true;
            if (i % primes[j] == 0) phi[i * primes[j]] = phi[i] * primes[j];
            else phi[i * primes[j]] = phi[i] * (primes[j] - 1);
        }
    }
}

int main()
{
    init();
    int n;
    while (scanf("%d", &n) != EOF) {
        LL ans = 0;
        --n;
        for (int i = 2; i <= n; ++i) phi[i] += phi[i - 1]; 
        ans += phi[n] * 2 + 1;
        printf("%lld
", ans);
    }
    return 0;
}