【BZOJ 2721】 樱花

【题目链接】

点击打开链接

【算法】

令n!=z,因为1 / x + 1 / y = 1 / z,所以x,y>z，不妨令y = z + d

则1 / x + 1 / (z + d) = 1 / z

1 / x = 1 / z - 1 / (z + d)

1 / x = d / (z + d)z

x = z(z + d) / d = z^2 / d + z

因为x是正整数，所以z^2 / d是正整数，所以d | z^2

问题就转化为了求z^2的约数个数

约数个数定理 x = p1^k1p2^k2....pn^kn,(p1,p2,....pn)为质数，x的约数个数为(k1+1)(k2+1)...(kn+1)

分解质因数即可

【代码】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e6;
const int MOD = 1e9 + ;

ll N,i,k,ans=,tot=;
ll prime[MAXN+],f[MAXN+],sum[MAXN+];

template <typename T> inline void read(T &x) {
        ll f=; x=;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c); c = getchar()) x = x *  + c - '';
        x *= f;
}

template <typename T> inline void write(T x) {
    if (x < ) { putchar('-'); x = -x; }
    if (x > ) write(x/);
    putchar(x%+'');
}

template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
}

inline void sieve(ll n) {
        ll i,j,tmp;
        for (i = ; i <= n; i++) {
                if (!f[i]) prime[++tot] = f[i] = i;
                for (j = ; j <= tot; j++) {
                        tmp = i * prime[j];
                        if (tmp > n) break;
                        f[tmp] = prime[j];
                        if (f[i] == prime[j]) break;
                }
        }
}

inline void calc(ll x) {
        while (x != ) {
                sum[f[x]]++;
                x /= f[x];
        }
}

int main() {

        read(N);
        sieve(N);
        for (i = ; i <= N; i++) calc(i);
        for (i = ; i <= tot; i++) ans = ans * ( * sum[prime[i]] + ) % MOD;
        writeln(ans);

        return ;

}