打表找规律: 当n为质数是,GCD(n)=n; 当n为质数k的q次方时,GCD(n)=k; 其他情况,GCD(n)=1. 代码如下: #include<iostream> #include<cstdlib> #include<stdio.h> #define ll long long #define M 1000001 using namespace std; ll a[M]; ],cnt; bool f[M]; int fac(int n) { ;i<cnt&a…

This time I need you to calculate the f(n) . (3<=n<=1000000) f(n)= Gcd(3)+Gcd(4)+-+Gcd(i)+-+Gcd(n). Gcd(n)=gcd(C[n][1],C[n][2],--,C[n][n-1]) C[n][k] means the number of way to choose k things from n some things. gcd(a,b) means the greatest common di…

定义$Gcd(n)=gcd(\binom{n}{1},\binom{n}{2}...\binom{n}{n-1})$,$f(n)=\sum_{i=3}^{n}{Gcd(i)}$,其中$(3<=n<=1000000)$. 由于组合数是二项式,Gcd()则是把首位两项去掉后所有项间进行gcd,那么我们可知当n为素数时,根据组合数公式,该素数不可能被其分母阶乘中的某个数除掉,那么每项都有该素数留下来,所以$Gcd(p) = p$,再推广,如果该数是某单个素数的幂指倍,那么同理仍然会有素数留下来所以$…

原题直通车:HDU 4734 F(x) 题意:F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 1, 求0.....B中F[x]<=F[A]的个数. 代码: // 31MS 548K 931 B G++ #include<iostream> #include<cstdio> #include<cstring> using namespace std; int digit[11], dp[11][6000],…

calculate the f(n) . (3<=n<=1000000)f(n)= Gcd(3)+Gcd(4)+-+Gcd(i)+-+Gcd(n).Gcd(n)=gcd(C[n][1],C[n][2],--,C[n][n-1])C[n][k] means the number of way to choose k things from n some things. 网络上有这个题的题解,但是都是说打表找规律,没有给出规律的证明.昨天睡前yy了一下,给个证明: 首先规律是: 1.Gcd(n)=…

Giving the N, can you tell me the answer of F(N)? Input Each test case contains a single integer N(1<=N<=10^9). The input is terminated by a set starting with N = 0. This set should not be processed. Output For each test case, output on a line the v…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=4734 数位DP. 用dp[i][j][k] 表示第i位用j时f(x)=k的时候的个数,然后需要预处理下小于k的和,然后就很容易想了 dp[i+1][j][k+(1<<i)]=dp[i][j1][k];(0<=j1<=j1) AC代码: #include <iostream> #include <cstdio> #include <cstring> #i…

题目链接 F(N) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4579 Accepted Submission(s): 1610 Problem Description Giving the N, can you tell me the answer of F(N)? Input Each test case contains a si…

题目链接:HDU 2009-4 Programming Contest 分析:具有一定的周期性——4018处理下就可以A了 Sample Input Sample Output AC代码: #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<iostream> #include<queue> #include<map>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4734 Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n…