题意:给你一个数,问这个数能否等于一系列连续的质数的和: 解题思路:质数筛打出质数表:然后就是尺取法解决: 代码: #include<iostream> #include<algorithm> #include<cstring> #define maxn 1000005 using namespace std; int visit[maxn];int prime[maxn]; void qprime() { memset(visit,,sizeof(visit)); ;…

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. The intege…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 23931   Accepted: 13044 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

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…

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…

题目大意 给定一个数字N,N可能由1个或多个连续的素数求和得到,比如41 = 2+3+5+7+11+13, 41 = 11+13+17, 41 = 41.求出对于N,所有可能的组合形式. 题目分析 先求出所有可能构成加数的素数,使用埃氏筛选法.然后求出所有的可能形式,由于所选择的是一个连续的区间,可以使用一个头指针,一个尾指针,区间选择为头尾指针内部的区域,通过头尾指针的移动来更改区间.即尺取法.     尾部保持不动,不断增加头部,并加上头部数据,记录区间内的和,若恰好等于n,则计数加1,若大…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22876   Accepted: 12509 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

poj3061 Subsequence 题目链接: http://poj.org/problem?id=3061 挑战P146.题意:给定长度为n的数列整数a0,a1,...,a(n-1)以及整数S,求出总和不小于S的连续子序列的长度的最小值,如果解不存在,则输出零.$10<n<10^5,0<a_i<=10^4,S<10^8$; 思路:尺取法,设起始下标s,截止下标e,和为sum,初始时s=e=0;若sum<S,将sum增加a(e),并将e加1,若sum>=S,更…

传送门 NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 1585    Accepted Submission(s): 688 Description NanoApe, the Retired Dog, has returned back to prepare for for the…

题目大意:从给定序列里找出区间和大于等于S的最小区间的长度. 前阵子在zzuli OJ上见过类似的题,还好当时补题了.尺取法O(n) 的复杂度过掉的.尺取法:从头遍历,如果不满足条件,则将尺子尾 部增加,若满足条件,则逐渐减少尺子头部直到不满足条件为止,保存 尺子长度的最小值(尾部-头部+1)即可. 理论上累计区间和+二分查找的暴力也能过. 代码如下: #include <stdio.h> #include <algorithm> #include <string.h>…