2820 YY的GCD
思路:
莫比乌斯反演+整除分块
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb emplace_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << "\n"; #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); //head const int N = 1e7 + 10; int T, n, m; int prime[N/10], mu[N], sum[N], cnt; bool not_p[N]; LL s[N]; void seive() { mu[1] = 1; for (int i = 2; i < N; ++i) { if(!not_p[i]) prime[++cnt] = i, mu[i] = -1; for (int j = 1; j <= cnt && i*prime[j] < N; ++j) { not_p[i*prime[j]] = true; if(i%prime[j] == 0) { mu[i*prime[j]] = 0; break; } mu[i*prime[j]] = -mu[i]; } } for (int i = 2; i < N; ++i) { if(!not_p[i]) for (int j = i; j < N; j += i) sum[j] += mu[j/i]; } for (int i = 1; i < N; ++i) s[i] = s[i-1] + sum[i]; } int main() { seive(); scanf("%d", &T); while(T--) { scanf("%d %d", &n, &m); LL ans = 0; int up = min(n, m); for (int l = 1, r; l <= up; l = r+1) { r = min(n/(n/l), m/(m/l)); ans += (n/l)*1LL*(m/l)*(s[r]-s[l-1]); } printf("%lld\n", ans); } return 0; }