题意: 给出n,求把n写成若干个连续素数之和的方案数. 分析: 这道题非常类似大白书P48的例21,上面详细讲了如何从一个O(n3)的算法优化到O(n2)再到O(nlogn),最后到O(n)的神一般的优化. 首先筛出10000以内的素数,放到一个数组中,然后求出素数的前缀和B.这样第i个素数一直累加到第j个素数,就可表示为Bj - Bi-1 枚举连续子序列的右端点j,我们要找到Bj - Bi-1 = n,也就是找到Bi-1 = Bj - n. 因为Bj是递增的,所以Bi-1也是递增的,所以我们就…
Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…
题目链接: 传送门 Sum of Consecutive Prime Numbers Time Limit: 1000MS     Memory Limit: 65536K Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive intege…
POJ2739 Sum of Consecutive Prime Numbers 题目大意:给出一个整数,如果有一段连续的素数之和等于该数,即满足要求,求出这种连续的素数的个数 水题:艾氏筛法打表+尺取法区间推进,0ms水过(注意循环的终止条件) #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include &l…
Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25225   Accepted: 13757 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…
Sum of Consecutive Prime Numbers http://poj.org/problem?id=2739 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28929   Accepted: 15525 Description Some positive integers can be represented by a sum of one or more consecutive prime numbe…
Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19697   Accepted: 10800 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…
POJ.2739 Sum of Consecutive Prime Numbers(水) 代码总览 #include <cstdio> #include <cstring> #include <vector> #include <cmath> #define nmax 10005 using namespace std; int prime[nmax],n; vector<int> vprime; void init() { memset(pri…
                                                                                                         Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24498   Accepted: 13326 Description Some positive i…
Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19895   Accepted: 10906 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…
解题思路:给定一个数,判定它由几个连续的素数构成,输出这样的种数 用的筛法素数打表 Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20020   Accepted: 10966 Description Some positive integers can be represented by a sum of one or more consecutive…
POJ 2739 Sum of Consecutive Prime Numbers(素数) http://poj.org/problem? id=2739 题意: 给你一个10000以内的自然数X.然后问你这个数x有多少种方式能由连续的素数相加得来? 分析: 首先用素数筛选法把10000以内的素数都找出来按从小到大保存到prime数组中. 然后找到数X在prime中的上界, 假设存在连续的素数之和==X, 那么一定是从一个比X小的素数開始求和(不会超过X的上界),直到和sum的值>=X为止. 所…
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22019 Accepted: 12051 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations d…
 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 5…
Time Limit: 1000MS   Memory Limit: 65536K Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two r…
Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53…
https://vjudge.net/problem/UVA-1210 统计质数前缀和,枚举左右端点,这一段的区间和+1 #include<cstdio> #define N 10001 using namespace std; int cnt,p[N],sum[N],ans[N]; bool v[N]; int main() { ;i<N;i++) { if(!v[i]) { v[i]=true; p[++cnt]=i; } ;j<=cnt;j++) { if(i*p[j]>…
题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3651 题意:问一个数n能拆成多少种连续的素数和. 筛出所有素数发现只有1200个,维护素数的前缀和,再打出所有区间的前缀和的表,统计出来,直接查询. #include <bits/stdc++.h> using namespace std; ; bool isprim…
 素数之和 题目大意:一些整数可以表示成一个连续素数之和,给定一个整数要你找出可以表示这一个整数的连续整数序列的个数 方法:打表,然后用游标卡尺法即可 #include <iostream> #include <functional> #include <algorithm> #define MAX_N 10010 using namespace std; static int primes_set[MAX_N], flag[MAX_N], p_sum; void in…
题目链接:http://poj.org/problem?id=2739 预处理出所有10001以内的素数,按照递增顺序存入数组prime[1...total].然后依次处理每个测试数据.采用双重循环计算n的表示数: 外循环i :  for (i = 0; x >= prime[i]; i++) 的循环结构枚举所有可能的最小素数prime[i]: 内循环:   while (ans < x && j < total)   ans += prime[j++];    计算连续…
简单的素数打表,然后枚举.开始没注意n读到0结束,TLE了回..下次再认真点.A过后讨论里面有个暴力打表过的,给跪了! #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> #include <cctype> #include <complex.h> #include <cmat…
题意:给一个数 可以写出多少种  连续素数的合 思路:直接线性筛 筛素数 暴力找就行   (素数到n/2就可以停下了,优化一个常数) 其中:线性筛的证明参考:https://blog.csdn.net/nk_test/article/details/46242401 https://blog.csdn.net/qq_40873884/article/details/79124552 https://blog.csdn.net/baoli1008/article/details/50788512…
参考:https://www.cnblogs.com/baozou/articles/4481191.html #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; bool prime[N]; void primemap() { memset(prime,true,sizeof(prime)); prime[]=prime[]=false; ;i<N;i++…
题目链接:http://poj.org/problem?id=2739 #include <cstdio> #include <cstring> using namespace std; int method[10001][1300]; int dp[10001]; bool isntprime[10001]; int heap[1300],cnt; void calprime(){ method[2][0]=1; dp[2]++; heap[cnt++]=2; for(int i…
题目大意 给定N,要求你计算用连续的素数的和能够组成N的种数 题解 先筛选出素数,然后暴力判断即可... 代码: #include<iostream> #include<cstring> using namespace std; #define MAXN 10000 ],cnt; ]; void get_prime() { cnt=; memset(check,false,sizeof(check)); ;i<=MAXN;i++) { if(!check[i]) prime[…
[题目链接] http://poj.org/problem?id=2739 [题目大意] 求出一个数能被拆分为相邻素数相加的种类 [题解] 将素数筛出到一个数组,题目转化为求区段和等于某数的次数,尺取法即可. [代码] #include <cstdio> #include <algorithm> using namespace std; int p[10010],cnt,a[10010],x; int main(){ for(int i=2;i<=10000;i++){ if…
给一个数 写成连续质数的和的形式,能写出多少种 *解法:先筛质数 然后尺取法 **尺取法:固定区间左.右端点为0,如果区间和比目标值大则右移左端点,比目标值小则右移右端点                 详见http://blog.csdn.net/consciousman/article/details/52348439 #include <iostream> #include <cstdio> using namespace std; #define SZ 11000 bool…
POJ 2739 Sum of Consecutive Prime Numbers Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu   Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representati…
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 27853 Accepted: 14968 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations d…
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…