luogu2257 YY的GCD--莫比烏斯反演
阿新 • • 發佈:2019-01-20
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給定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)為質數的(x, y)有多少對
多組數據T = 10000
N, M <= 10000000
推式子
\(\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=p]\)
\(=\sum_p\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}[\gcd(i,j)=1]\)
\(=\sum_p\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}\sum_{d|i,d|j}\mu(d)\)
\(=\sum_{d=1}^n\mu(d)\sum_p\lfloor\frac n{dp}\rfloor\lfloor\frac m{dp}\rfloor\)
令\(q=dp\)
\(=\sum_{q=1}^n(\sum_{p|q}\mu(\frac q p))\lfloor\frac nq\rfloor\lfloor\frac mq\rfloor\)
\(\mu\)線性篩
然後在對於質數枚舉倍數求對於每個\(i\)的\(\sum_{p|i}\mu(\frac i p)\)
然後打數論分塊就行了
#include <cstdio> #include <functional> using namespace std; const int fuck = 10000000; int prime[10000010], tot; bool vis[10000010]; int mu[10000010], sum[10000010]; int main() { mu[1] = 1; for (int i = 2; i <= fuck; i++) { if (vis[i] == false) prime[++tot] = i, mu[i] = -1; for (int j = 1; j <= tot && i * prime[j] <= fuck; j++) { vis[i * prime[j]] = true; if (i % prime[j] == 0) break; mu[i * prime[j]] = -mu[i]; } } for (int i = 1; i <= tot; i++) for (int j = 1; j * prime[i] <= fuck; j++) sum[j * prime[i]] += mu[j]; for (int i = 1; i <= fuck; i++) sum[i] += sum[i - 1]; int t; scanf("%d", &t); while (t --> 0) { int n, m; long long ans = 0; //別忘了初始化。。。 scanf("%d%d", &n, &m); if (n > m) {int t = m; m = n; n = t; } for (int i = 1, j; i <= n; i = j + 1) { j = min(n / (n / i), m / (m / i)); ans += (sum[j] - sum[i - 1]) * (long long)(n / i) * (m / i); } printf("%lld\n", ans); } return 0; }
luogu2257 YY的GCD--莫比烏斯反演