素数判断: 一.根据素数定义,该数除了1和它本身以外不再有其他的因数. 详见代码. int prime() { ; i*i<=n; i++) { ) //不是素数 ; //返回1 } ; //是素数返回0 } 二.打表,将所有的素数一一列出,存在一个数组里. 详见代码. void prime() { ; i<; i++) //从2开始一个一个找 { ) //这一个判断可以减少很多重复的,节省很多时间 { ; i*j<; j++) //只要乘以i就一定不是素数 { hash[i*j]=;…

看懂: Max Factor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3089    Accepted Submission(s): 985 Problem Description To improve the organization of his farm, Farmer John labels each of his N…

Max Factor Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description To improve the organization of his farm, Farmer John labels each of his N (1 <= N <= 5,000) cows with a distinct serial number in the ran…

题目描述: 题目大意:找出具有最大素数因子的整数.如果有不止一个,则输出在输入文件中出现最早的一个. 解题思路:刚开始时,p数组中的元素全为0,刚开始对于素数 i,p[i]=0,用一个for循环,将是素数 i 的倍数的数 的在数组p中的值全部赋值为 i: 如:第一轮:2为素数,p[2]=0,p[4]=2,p[6]=2,p[8]=2......p[2*n]=2; 第二轮:3为素数,  p[3]=0,p[6]=3,p[9]=3,p[12]=3......p[3*n]=3; 第...轮:i为素数,p[…

题目 //下面这个是最先用的方法,因为学姐先讲完这个,所以懒得写代码,就将就着这个用,结果搞了老半天,还是错了,心累.. #include<stdio.h> #include<string.h> int prime[5010]; bool isprime[5010]; int solve(int n) { memset(prime,0,sizeof(prime)); int p=0; for(int i=0; i<=n; i++) isprime[i]=true; ispri…

Max Factor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10245    Accepted Submission(s): 3304 Problem Description To improve the organization of his farm, Farmer John labels each of his N (1…

Problem Description Everybody knows any number can be combined by the prime number. Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. Specially, LPF(1) = 0.   In…

题目链接:传送门 题目: Prime Distance Time Limit: 1000MS Memory Limit: 65536K Total Submissions: Accepted: Description The branch of mathematics called number theory and itself). The first prime numbers are ,,, but they quickly become less frequent. One of the…

用素数筛选法即可. 范围long int,其实大小范围和int一样,一开始以为是指long long,想这就麻烦了该怎么弄. 而现在其实就是int的范围,那难度档次就不一样了,瞬间变成水题一枚,因为int根号后大小不超过60000. 即因子的大小不会超过60000,除非本身是质数. int 4 -2147438648-+2147438647long int 4 -2147438648-+2141438647 #include <iostream> #include <cstdio>…

题目大意: 有一个数n,满足lcm(i,j)==n并且i<=j时,(i,j)有多少种情况? 解题思路: n可以表示为:n=p1^x1*p2^x1.....pk^xk. 假设lcm(a,b) == n: a = p1^c1 * p2^c2 ..... pk^ck. b = p1^e1 * p2^e2 .... pk^ek. xi = max(ci, ei). 对于有序数对(a,b),有唯一分解定理知,每一个素因数的幂都决定了一个独一无二的数. 求(a,b)的种数就可以转化为求(ci,ei)的种数:…