不是很难，这里筛法写错了，一度以为自己推错了。。
题意：求解$ans = sum_{x = 1}^{a}sum_{y = 1}^{b}|gcd(x,y) = d|$
Solution:
$ans = = sum_{x = 1}^{[frac{a}{d}]}sum_{y = 1}^{[frac{b}{d}]}|gcd(x,y) = 1| = sum_{t = 1}^{min(frac{a}{d},frac{b}{d})} mu (t) * [frac{a}{dt}] * [frac{b}{dt}]$
其实化到了这里已经可以了，但我一开始继续化了，令T = dt.
那么$ans = sum_{T = 1}^{min(a,b)}mu (frac{T}{d}) * [frac{a}{T}] * [frac{b}{T}]$
然后就可以整除分块，然后处理一个莫比乌斯的前缀和，注意这里的莫比乌斯函数下标是T/d，分块时注意位置即可。
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<double,int> pii; const int N = 5e4+5; const int M = 1e6+5; const LL Mod = 1e9+7; #define rg register #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline int read() { int x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();} return x * f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; int mu[N],prime[N],tot = 0; LL sum[N]; bool vis[N]; void init() { mu[1] = 1; for(rg int i = 2;i < N;++i) { if(!vis[i]) { prime[++tot] = i; mu[i] = -1; } for(rg int j = 1;j <= tot && prime[j] * i < N;++j) { vis[prime[j] * i] = 1; if(i % prime[j] == 0) break; else mu[prime[j] * i] = -mu[i]; } } for(rg int i = 1;i < N;++i) sum[i] = sum[i - 1] + mu[i]; } int main() { init(); int ca;ca = read(); while(ca--) { int a,b,d;a = read(),b = read(),d = read(); LL ans = 0; for(int L = 1,r = 0;L <= min(a,b);L = r + 1) { r = min(a / (a / L),b / (b / L)); ans += (sum[r / d] - sum[(L - 1) / d]) * (a / L) * (b / L); } printf("%lld ",ans); } //system("pause"); return 0; }
现在我们直接看第一种，其实只要把后面的看成$frac{frac{a}{d}}{t} $和$frac{frac{b}{d}}{t} $来分块即可。
#include<bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<double,int> pii; const int N = 5e4+5; const int M = 1e6+5; const LL Mod = 1e9+7; #define rg register #define pi acos(-1) #define INF 1e18 #define CT0 cin.tie(0),cout.tie(0) #define IO ios::sync_with_stdio(false) #define dbg(ax) cout << "now this num is " << ax << endl; namespace FASTIO{ inline int read() { int x = 0,f = 1;char c = getchar(); while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();} while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();} return x * f; } void print(int x){ if(x < 0){x = -x;putchar('-');} if(x > 9) print(x/10); putchar(x%10+'0'); } } using namespace FASTIO; int mu[N],prime[N],tot = 0; LL sum[N]; bool vis[N]; void init() { mu[1] = 1; for(rg int i = 2;i < N;++i) { if(!vis[i]) { prime[++tot] = i; mu[i] = -1; } for(rg int j = 1;j <= tot && prime[j] * i < N;++j) { vis[prime[j] * i] = 1; if(i % prime[j] == 0) break; else mu[prime[j] * i] = -mu[i]; } } for(rg int i = 1;i < N;++i) sum[i] = sum[i - 1] + mu[i]; } int main() { init(); int ca;ca = read(); while(ca--) { int a,b,d;a = read(),b = read(),d = read(); LL ans = 0; a = a / d,b = b / d; for(int L = 1,r = 0;L <= min(a,b);L = r + 1) { r = min(a / (a / L),b / (b / L)); ans += (sum[r] - sum[L - 1]) * (a / L) * (b / L); } printf("%lld ",ans); } //system("pause"); return 0; }