标签:ace pre src idt mes ++i 中间 algo 代码
题意:设a, b, c, d, k。 x属于[a, b], y属于[c, d]。问满足gcd(x, y)=k的(x, y)的对数是多少?注意:a=c=1;
公式推导:
注意一下,中间过程别爆精度:
ac代码:
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define ll long long const int N = 1e5 + 5; bool vis[N]; int mu[N], prime[N]; void mobiws() { int cnt = 0; mu[1] = 1; for (int i = 2; i < N; ++i) { if (!vis[i]) { prime[++cnt] = i; mu[i] = -1; } for (int j = 1; j<=cnt&&prime[j] * i < N; ++j) { vis[prime[j] * i] = 1; if (i%prime[j] == 0){ mu[prime[j] * i] = 0; break; } mu[prime[j] * i] =- mu[i]; } } for (int i = 2; i < N; ++i)mu[i] += mu[i - 1]; } ll cal(int n, int m, int d) { if (!n || !m || !d)return 0; n /= d; m /= d; int last; ll ans = 0; for (int i = 1; i <= min(n, m); i = last + 1) { last = min(n / (n / i), m / (m / i)); ans += (mu[last] - mu[i - 1])*(1LL)*(n / i)*(1LL)*(m / i); } return ans; } int main() { mobiws(); int t, a, b, c, d, k, cas = 0; scanf("%d", &t); while (t--) { scanf("%d%d%d%d%d", &a, &b, &c, &d, &k); printf("Case %d: ", ++cas); ll ans = cal(b, d, k) - cal(min(b, d), min(b, d), k)/2; printf("%lld\n", ans); } }
标签:ace pre src idt mes ++i 中间 algo 代码
原文地址:https://www.cnblogs.com/ALINGMAOMAO/p/9683720.html