LOJ6241 性能优化

题意

T=1in1jni1kj[gcd(j,k)=1]T = \sum_{1\le i\le n}\sum_{1\le j\le\lfloor \frac{n}{i} \rfloor}\sum_{1\le k\le j}[\gcd(j,k)=1] 多组数据 n10106n\le 10^{10^6}

题解

T=1in1jniφ(j)T = \sum_{1\le i\le n}\sum_{1\le j\le\lfloor \frac{n}{i} \rfloor}\varphi(j)S(n)=1inφ(i)\text{S}(n) = \sum_{1\le i\le n}\varphi(i) 则有 T=1inS(ni)=S(n)+2inS(ni)\begin{aligned} T &= \sum_{1\le i\le n}\text{S}(\lfloor \frac{n}{i} \rfloor) \\ &= \text{S}(n) + \sum_{2\le i\le n}\text{S}(\lfloor \frac{n}{i} \rfloor) \\ \end{aligned} 根据杜教筛,由 φI=Id\varphi*\text{I}=\text{Id},有 I(1)×S(n)=1inId(i)2inS(ni)×I(i) \text{I}(1)\times \text{S}(n)=\sum_{1\le i\le n}\text{Id}(i)-\sum_{2\le i\le n}\text{S}(\lfloor\frac{n}{i}\rfloor)\times \text{I}(i) S(n)=n(n+1)22inS(ni) \text{S}(n)=\frac{n(n+1)}{2}-\sum_{2\le i\le n}\text{S}(\lfloor\frac{n}{i}\rfloor) 带回原式得 T=n(n+1)22inS(ni)+2inS(ni)=n(n+1)2 \begin{aligned} T=&\frac{n(n+1)}{2}-\sum_{2\le i\le n}\text{S}(\lfloor\frac{n}{i}\rfloor) + \sum_{2\le i\le n}\text{S}(\lfloor \frac{n}{i} \rfloor)\\ =&\frac{n(n+1)}{2} \end{aligned}

作者

Fisher Cai

发布于

2021-01-15

更新于

2025-07-22

许可协议

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