Codeforces1062B. Math(合数分解)】的更多相关文章

Cards Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 470    Accepted Submission(s): 72 Problem Description Given some cards each assigned a number, you're required to select EXACTLY K cards amo…
GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 40    Accepted Submission(s): 22 Problem Description Given two positive integers G and L, could you tell me how many solutions of (x,…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4497 GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2151    Accepted Submission(s): 955 Problem Description Given two positive integer…
RGCDQ Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2818    Accepted Submission(s): 1108 Problem Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and m…
Rabbit Kingdom Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1999    Accepted Submission(s): 689 Problem Description Long long ago, there was an ancient rabbit kingdom in the forest. Every ra…
题意:x可以表示为bp, 求这个p的最大值,比如 25=52, 64=26,  然后输入x 输出 p 就是一个质因子分解.算法.(表示数据上卡了2个小时.) 合数质因子分解模板. ]; ]; ; ;n!=;++i) { ) { num[cnt]=i; ){ind[cnt]++;n/=i;} cnt++; } )break; } ){num[cnt]=n; ind[cnt++]=;} 两种方法: 方法一:时间最坏的时间复杂度是(大概10^8*n)就是这种方法,数据卡了很久,如果数据再给狠一点肯定不…
题意:给你一个个数对a, b 表示ab这样的每个数相乘的一个数n,求n-1的质数因子并且每个指数因子k所对应的次数 h. 先把合数分解模板乖乖放上: ; ans != ; ++i) { ) { num[cnt] = i; ; ){ ++k; ans /= i; } index[cnt++] = k; } )break; } ){ num[cnt] = ans; index[cnt++] = ; } 然后,我自己写了个快速幂 快速幂的模板: ll pow(ll a, ll n) { ll res;…
题目链接:传送门 题目: B. Math time limit per test second memory limit per test megabytes input standard input output standard output JATC's math teacher always gives the class some interesting math problems so that they don't get bored. Today the problem is a…
就是输出n时,莫比乌斯函数的值.直接将n唯一分解即可. 思路:筛出105以内的素数,因为109开方,105就差不多.当一个大数还没有被1000个素数分解,那么这个数基本上可以认为是素数(为合数为小概率了).使用欧拉筛筛出105以内的素数,然后枚举素数试除即可 ac代码: #include<cstdio> #include<cstring> #define maxn int(1e5) #define ll long long int prime[maxn]; bool vis[max…
1801: Mr. S’s Romance Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 15  Solved: 5[Submit][Status][Web Board] Description Mr. S is a young math professor who is famous for being crazy about collecting prime numbers. Once he met a number, he would fir…