Largest prime factor Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13481    Accepted Submission(s): 4765 Problem Description Everybody knows any number can be combined by the prime number.Now,…

题意:给你一个数,让你求它的最大因子在素数表的位置. 析:看起来挺简单的题,可是我却WA了一晚上,后来终于明白了,这个第一层循环不是到平方根, 这个题和判断素数不一样,只要明白了这一点,就很简单了. 代码如下: #include <iostream> #include <cstdio> #include <algorithm> #include <queue> #include <vector> #include <cstring>…

题目梗概:求1000000以内任意数的最大质因数是第几个素数,其中 定义 1为第0个,2为第1个,以此类推. #include<string.h> #include<stdio.h> #include<math.h> ],b[],c[];//b[i]表示i是第几个素数,c[k]表示k的最大素数是c[k],a[i]表示是不是素数 int main() { int n,i,num,j; memset(a,,sizeof(a)); a[]=; b[]=; num=; ;i&l…

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. Inpu…

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.   Input…

#include <iostream> #include <vector> #include <cmath> using namespace std; const int MAX=1000001; bool isPrime[MAX];//isPrime[i]=true表示是素数,false表示不是 int indexes[MAX]; //存放素数因子的序号 void eraosthenes() { fill(isPrime,isPrime+MAX,true);//如果0…

题目大意:求出比给出数小的互质的质数个数. 题解:直接用筛法求素数,稍微改编一下,将原先的布尔数组变为数组用来记录信息就可以了. 注意点:大的数组定义要放在程序的开头,不要放在main里面,不然会栈溢出. #include <cstdio> #define max 1000000 }; int main() { int n; ; ; i<max; ++i) { if(prim[i]) continue; for(int j=i; j<max; j+=i) prim[j]=cnt;…

传送门 题意 给出若干个数n(n<=1000000),求每个n的最大质因子的排名. 质数的排名:如果素数p是第k小的素数,那么p的排名就是k. 思路 乍一看不知道怎么搞. 其实可以想想我们怎么筛素数的,每个数都会被它的质因数筛去. 这就和题目一样了. 代码 #include <cstdio> ; ; int main() { int i, j, x; ; i < MAXN; i++) if(!notpri[i]) { for(j = i; j < MAXN; j += i)…

Largest prime factor Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6990    Accepted Submission(s): 2471 Problem Description Everybody knows any number can be combined by the prime number. Now…

/*打表把素数能组合的数先设置成相应的位数*/ /* if n equals two and n is No.1 position of prime factors  so four position of prime factors is no.1,as well*/ /* although two can combined six but three also can combined six */ /* and three position of prime factors is two…