题意
求$gcd(i, j)$为素数的对数，$ i leqslant N ,j leqslant N$
Sol
一开以为要用zap那题的思路暴力求，但是化到一半发现会做了qwq。
设$p_i$为第$i$个素数
我们要求的是
$$ans = sum_{i = 1}^n sum_{j = 1}^n [gcd(i, j) = p_i]$$
$$ = sum_{i = 1}^{frac{n}{p_i}} sum_{j = 1}^{frac{n}{p_i}} [gcd(i, j) = 1]$$
根据定理
$$sum_{i = 1}^n sum_{j = 1}^n [gcd(i, j) = 1] = 2phi(i) - 1$$
而且$10^7$以内的质数只有大约$6 * 10^5$个
直接对$phi$做前缀和然后暴力算就行了
// luogu-judger-enable-o2 #include<cstdio> #include<algorithm> #include<cmath> #define LL long long #define rg register //#define int long long using namespace std; const int MAXN = 1e7 + 10, mod = 1e9 + 7; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int prime[MAXN], vis[MAXN], tot; LL phi[MAXN]; void GetPhi(int N) { vis[1] = phi[1] = 1; for(rg int i = 2; i <= N; i++) { if(!vis[i]) prime[++tot] = i, phi[i] = i - 1; for(rg int j = 1; j <= tot && i * prime[j] <= N; j++) { vis[i * prime[j]] = 1; if(!(i % prime[j])) {phi[i * prime[j]] = phi[i] * prime[j]; break;} phi[i * prime[j]] = phi[i] * phi[prime[j]]; } } for(rg int i = 2; i <= N; i++) phi[i] += phi[i - 1]; } inline LL F(int N) { return (phi[N] << 1) - 1; } int main() { GetPhi(1e7 + 5); int N = read(); LL ans = 0; for(rg int i = 1; i <= tot && prime[i] <= N; i++) ans += F(N / prime[i]); printf("%lldn", ans); return 0; } /* */
内容来源于网络如有侵权请私信删除
- 还没有人评论，欢迎说说您的想法！
客服
