C - Primes and Multiplication 思路:找到x的所有质数因子,用一个vector储存起来,然后对于每一个质因子来说,我们要找到它对最后的答案的贡献的大小,即要找到它在最后的乘积中出现了多少次. 求解方法: for(auto i:v) { ll cnt=0; ll temp=n/i; while(temp) cnt+=temp,temp/=i; ans=(ans*qpow(i,cnt))%mod; } 另外,该题还需要用到快速幂的技巧. ll qpow(ll x,ll n…

题目链接:https://www.luogu.org/problem/CF1228C 问题可以转化为:求质数 $p$ 在 $1\sim n$ 中的每个数中的次幂之和. 因为 $p$ 是一个质数,只能由 $1$ 乘以 $p$ 表示出来,所以可以将问题转化为求 $p$ 在 $n!$ 中出现的次幂. 我们可以像提取公因式一样地去提取这个 $p$. 那么,先考虑 $p$ 的贡献:$1\sim n$ 中能被 $p$ 整除的乘积为 $p^{\frac{n}{p}}\times (\frac{n}{p}!)$…

链接: https://codeforces.com/contest/1228/problem/C 题意: Let's introduce some definitions that will be needed later. Let prime(x) be the set of prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}. Let g(x,p) be the maximum possible int…

传送门 当然是考虑 $n$ 的每个质数 $p$ 对答案的贡献 考虑 $p^k$ 在 $[1,m]$ 中出现了几次,显然是 $\left \lfloor \frac{m}{p^k} \right \rfloor$ 次 那么对于 $p^k$ ,它目前的贡献就是 $p^{\left \lfloor \frac{m}{p^k} \right \rfloor}$ ,注意这里不是 $p^{k\left \lfloor \frac{m}{p^k} \right \rfloor}$,因为之后计算对于 $k'<k…

题目链接:http://codeforces.com/contest/1228/problem/C 题解:给定一个函数f,g,题目有描述其中的表达式含义和两者之间的关系. 然后计算: 首先把给定的x用唯一分解定理分解出素因子 因为在1-n中,n/p(素因子)的值就是其1-n中有多少个数能整除p,n/p^2就是有多少个数能被p^2整除,所以做一次循环,每次n=n/p,直到n为0,sum记录每次n除以p的个数,就是1-n中能整除p的次方的数的幂指数之和了. 举个例子,45和3,n=45,p=3,第一…

题目连接:https://codeforces.com/contest/1228/problem/C 题目大意:g(x,y)==y^k(其中y^k是X的最大可以整除因子) f(x,y)==g(x,p1)*(x,p2)....其中pi是x的第i个质因子 求f(x,1)*f(x,2)*f(x,3)---*f(x,n); 题解:对于整数N....N/x==从1到N可以被X整除的个数N/x^2同理... 所以我们不妨对每个质因子进行叠加,对于第一个p1,让N多次除以p1,同时记录可以整除的个数,最终我们…

目录 Contest Info Solutions A. Distinct Digits B. Filling the Grid C. Primes and Multiplication D. Complete Tripartite E. Another Filling the Grid Contest Info Practice Link Solved A B C D E F 5/6 O O O O Ø - O 在比赛中通过 Ø 赛后通过 ! 尝试了但是失败了 - 没有尝试 Solutions…

https://codeforces.com/contest/1228/problem/A A. Distinct Digits 超级简单嘻嘻,给你一个l和r然后寻找一个数,这个数要满足的条件是它的每一位的数字不相同,找出满足要求的最小的那个数输出,没有找到就输出-1: #include<bits/stdc++.h> using namespace std; bool check(int n){ ]={}; while(n){ ]) vis[n%]=; else return false; n…

4491. Primes in GCD Table Problem code: PGCD Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has cre…

SPOJ4491. Primes in GCD Table Problem code: PGCD Johnny has created a table which encodes the results of some operation -- a function of two arguments. But instead of a boring multiplication table of the sort you learn by heart at prep-school, he has…

Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits:Special thanks to @mithmatt for adding this problem and creating all test…

A multiplication game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6028   Accepted: 3013 Description Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p =…

                             Matrix Multiplication Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18173   Accepted: 3912 Description You are given three n × n matrices A, B and C. Does the equation A × B = C hold true? Input The first l…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the product of them. bobo hates big integers. So you are only asked to find t…

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. 译文: 10以下的素数之和为17,求出2000000以下的素数之和. ======================= 第一次code: import java.util.Scanner; public class Main { public static void main(String[]…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 820    Accepted Submission(s): 328 Problem Description Given two matrices A and B of size n×n, find the product of them. b…

A typical implementation Booth's algorithm can be implemented by repeatedly adding (with ordinary unsigned binary addition) one of two predetermined values A and S to a product P, then performing a rightward arithmetic shift on P. Let m and r be the…

题目大家都非常熟悉,求小于n的所有素数的个数. 自己写的python 代码老是通不过时间门槛,无奈去看了看大家写的code.下面是我看到的投票最高的code: class Solution: # @param {integer} n # @return {integer} def countPrimes(self, n): if n < 3: return 0 primes = [True] * n primes[0] = primes[1] = False for i in range(2,…

http://acm.hdu.edu.cn/showproblem.php?pid=4951 2014多校 第八题 1008 2014 Multi-University Training Contest 8 Multiplication table Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 438    Accepted Submi…

hdu4920 Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 568    Accepted Submission(s): 225 Problem Description Given two matrices A and B of size n×n, find the product o…

题目链接:http://poj.org/problem?id=1651 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the prod…

examination questions Description: Count the number of prime numbers less than a non-negative number, n References: How Many Primes Are There? Sieve of Eratosthenes Please use the following function to solve the problem: int countPrimes(int n){ } 解题代…

Count Primes Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits: Special thanks to @mithmatt for adding this problem and cre…

Count the number of prime numbers less than a non-negative number, n public int countPrimes(int n) { ArrayList<Integer> primes = new ArrayList<Integer>(); ) , n - ); primes.add(); primes.add(); ; i <= n; i++) { boolean isPrime = true; for (…

题目传送门 /* 题意:加上适当的括号,改变计算顺序使得总的计算次数最少 矩阵连乘积问题,DP解决:状态转移方程: dp[i][j] = min (dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j]) (i<=k<j) s[i][j] 记录断开的地方(即加括号的位置),回溯法输出结果 */ #include <cstdio> #include <cstring> #include <string> #include &l…

Matrix Chain Multiplication Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 834    Accepted Submission(s): 570 Problem DescriptionMatrix multiplication problem is a typical example of dynamical…

http://acm.hdu.edu.cn/showproblem.php?pid=4715 Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description All you know Goldbach conjecture.That is to say, Every even integer great…

Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3339    Accepted Submission(s): 953 Problem Description All you know Goldbach conjecture.That is to say, Every even int…

D - Matrix Multiplication Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description       Let us consider undirected graph G = {V; E} which has N vertices and M edges. Incidence matrix of this g…

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1775    Accepted Submission(s): 796 Problem Description Given two matrices A and B of size n×n, find the product of them.…