How many prime numbers Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12955    Accepted Submission(s): 4490 Problem Description   Give you a lot of positive integers, just to find out how many…

Problem Description Give you a lot of positive integers, just to find out how many prime numbers there are. Input There are a lot of cases. In each case, there is an integer N representing the number of integers to find. Each integer won't exceed 32-…

米勒罗宾素数测试: /* if n < 1,373,653, it is enough to test a = 2 and 3. if n < 9,080,191, it is enough to test a = 31 and 73. if n < 4,759,123,141, it is enough to test a = 2, 7, and 61. if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7,…

题意:给定一个数,判断是不是素数. 析:由于数太多,并且太大了,所以以前的方法都不适合,要用米勒拉宾算法. 代码如下: #include <iostream> #include <cstdio> #include <algorithm> #include <queue> #include <vector> #include <cstring> #include <map> #include <cctype> u…

Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 19350 Accepted: 10619 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations d…

POJ:3006 很显然这是一题有关于素数的题目. 注意数据的范围,爆搜超时无误. 这里要用到筛选法求素数. 筛选法求素数的大概思路是: 如果a这个数是一个质数,则n*a不是质数. 用一个数组实现就是: memset(prime,true,sizeof(prime)); if (prime[i]) prime[i*j]=false; 部分程序如下:(朴素) ; ]; memset(prime,true,sizeof(prime)); ; i <= ::max ; i ++ ) { ; j <=…

传送门:HDU 5895 Mathematician QSC 这是一篇很好的题解,我想讲的他基本都讲了http://blog.csdn.net/queuelovestack/article/details/52577212 [分析]一开始想简单了,对于a^x mod p这种形式的直接用欧拉定理的数论定理降幂了 结果可想而知,肯定错,因为题目并没有保证gcd(x,s+1)=1,而欧拉定理的数论定理是明确规定的 所以得另谋出路 那么网上提供了一种指数循环节降幂的方法 具体证明可以自行从网上找一找 有…

http://acm.hdu.edu.cn/showproblem.php?pid=2138 题意:给n个数判断有几个素数.(每个数<=2^32) #include <cstdio> using namespace std; typedef long long ll; ll ipow(ll a, ll b, ll m) { ll x=1; for(; b; b>>=1, (a*=a)%=m) if(b&1) (x*=a)%=m; return x; } ll rand…

题意: 给一个正整数N,找最小的M,使得N可以整除M,且N/M是质数. 数据范围: There are multiple test cases (no more than 1,000). Each case contains only one positive integer N.N≤1,000,000,000.Number of cases with N>1,000,000 is no more than 100. 思路: N=M*prime     故必有M或prime小于等于sqrt(N)…

数论题,本质是求出n的最大质因子 #include<time.h> #include <cstdio> #include <iostream> #include<algorithm> #include<math.h> #include <string.h> #include<vector> #include<queue> using namespace std; int main() { int n,t; wh…