51nod 1223 分数等式的数量

原题链接.

题解：

Ans=∑ni=1∑i−1j=1i+j | i∗j
=∑nd=1∑n/di=1∑i−1j=1d(i+j) | i∗j∗d2(gcd(i,j)=1)
=∑nd=1∑n/di=1∑i−1j=1i+j | i∗j∗d(gcd(i,j)=1)
∵gcd(i,j)=1
∴gcd(i+j,i)=gcd(i+j,j)=1
∴gcd(i+j,i∗j)=1
∴原式
=∑nd=1∑n/di=1∑i−1j=1i+j | d(gcd(i,j)=1)
=∑ni=1∑i−1j=1[gcd(i,j)=1]∗∑n/id=1[i+j | d]
=∑ni=1∑i−1j=1[gcd(i,j)=1]∗⌊n/ii+j⌋
=∑ni=1∑i−1j=1⌊n/ii+j⌋∑d|gcd(i,j)μ(d)
=∑nd=1∑n/di=1∑i−1j=1⌊n/i/dd(i+j)⌋
=∑nd=1∑n/di=1∑i−1j=1⌊nd2∗i∗(i+j)⌋
=∑nd=1∑n/di=1∑2i−1j=i+1⌊nd2∗i∗j⌋
=∑n√d=1∑n/d/d√i=1∑2i−1j=i+1⌊nd2∗i∗j⌋

暴力枚举d，i,分块j，是可以过的。
复杂度不会证明。

Code：

#include<cmath>#include<cstdio>#define ll long long#define fo(i, x, y) for(ll i = x; i <= y; i ++)#define min(a, b) ((a) < (b) ? (a) : (b))using namespace std;const int N = 320000;bool bz[N + 1]; int mu[N + 1], p[N];ll n, ans;int main() {    mu[1] = 1;    fo(i, 2, N) {        if(!bz[i]) p[++ p[0]] = i, mu[i] = -1;        fo(j, 1, p[0]) {            int k = i * p[j];            if(k > N) break;            bz[k] = 1;            if(i % p[j] == 0) {                mu[k] = 0; break;            }            mu[k] = -mu[i];        }    }    scanf("%lld", &n);    fo(d, 1, sqrt(n * 1.0)) if(mu[d]){        ll m = n / d / d;        fo(j, 2, sqrt(m * 1.0)) {            ll z = m / j;            fo(i, j + 1, 2 * j - 1) {                if(z / i == 0) break;                ll k = min(z / (z / i), 2 * j - 1);                ans += (z / i) * (k - i + 1) * mu[d];                i = k;            }        }    }    printf("%lld", ans);}