退役ACMer

性质：对于正整数n ∑d|nϕ(d)=n\sum_{d|n}\phi(d) = n 证明过程 (1)如果 n = 1 ϕ(n)=1=n\phi(n) = 1 = n 满足 (2)如果n是质数 ϕ(n)=n(1−1n)=n−1\phi(n) = n(1 - \frac{1}{n}) = n - 1 所以ϕ(n)+ϕ(1)=n\phi(n) + \phi(1) = n 满足 (3

传送门:UESTC 811这题思路出的还算快, 但是UESTC OJ评测卡了一天, 代码细节比较多, 调bug的时候都怀疑解法是不是又错了这题好像不好找题解题意1<=N<=10^10, 1<=K<=5求∑Ni=1∑Nj=1gcd(i,j)k\sum_{i=1}^N\sum_{j=1}^Ngcd(i, j)^k题解枚举gcd, 推下式子 ans=∑i=1N∑j=1Ngcd(i,j)k=2∑i=1N

传送门：51nod 1238题意求G(N)=∑i−1N∑j=1Nlcm(i,j) G(N) = \sum_{i - 1}^N\sum_{j=1}^Nlcm(i, j)题解首先G(N)=∑i=1N∑j=1Nlcm(i,j)=2∑i=1N∑j=1ilcm(i,j)−∑i=1Nlcm(i,i)=2∑i=1Ni∑d|i∑u=1idu[gcd(u,id)=1]−N(N+1)2=2∑i=1Ni∑d|iidϕ

题解参考1238code:#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1000001;const ll mod = 1e9 + 7;const ll inv = (mod + 1) / 2;const int mo = 2333333;bool isPrime[N];ll p

题解: 主要是原根的幂表示[1, p - 1], 然后转换 hint里的公式: 找了一篇证明:http://blog.youkuaiyun.com/Clove_unique/article/details/53152473代码:#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef l

杜教筛#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1000001;const ll mod = 1e9 + 7;const ll inv = (mod + 1) / 2;const ll _6 = (mod + 1) / 6;const int mo = 2333333;bo

题解a1moda0=ya_1 mod a_0 = y a2moda0=x∗y+ya_2 mod a_0 = x * y + y ….anmoda0=∑i=0n−1xi∗y=y∗(xn−1)/(x−1)a_n mod a_0 = \sum_{i = 0}^{n - 1} x^{i} * y = y * (x ^ n - 1) / (x - 1) 若y *(x - 1) % a0 =

题意:定以f(n)=(∑i=1nni)%(n+1)f(n) = (\sum_{i = 1} ^ nn ^ i) \% (n + 1)求g(n)=∑ni=1f(i)g(n) = \sum_{i = 1} ^ n f(i) n <= 1000000分析: 设n的质因子分解为 n=∏ki=1paiin=\prod_{i = 1}^kp_i^{a_i} 那么n、n^2…..n^n的表示: ni

传送门：SPOJ -NUMTRYE题解:首先∑ni=1gcd(n,i)=∑d|ndϕ(nd)\sum_{i=1}^ngcd(n, i) = \sum_{d|n}d\phi(\frac{n}{d}) 所以g(n)=∑ni=1n/gcd(n,i)=∑d|nndϕ(nd)=∑d|ndϕ(d)g(n) = \sum_{i=1}^nn/gcd(n, i) = \sum_{d|n}\frac{n}{d}\p

题解:设n=∏ni=1paiin = \prod_{i=1}^np_i^{a_i} 则ϕ(n)=∏ni=1paii(1−1pi)\phi(n)=\prod_{i=1}^np_i^{a_i}(1-\frac{1}{p_i})nϕ(n)=∏ni=11+1pi−1\frac{n}{\phi(n)}=\prod_{i=1}^n1 + \frac{1}{p_i-1}因为对于两个不同质数a,b(a < b)