描述 http://poj.org/problem?id=2739 多次询问,对于一个给定的n,求有多少组连续的素数,满足连续素数之和为n. Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22737   Accepted: 12432 Description Some positive integers can be represented by a su…

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…

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…

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…

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…

尺取法:顾名思义就是像尺子一样一段一段去取,保存每次的选取区间的左右端点.然后一直推进 解决问题的思路: 先移动右端点 ,右端点推进的时候一般是加 然后推进左端点,左端点一般是减 poj 2566 题意:从数列中找出连续序列,使得和的绝对值与目标数之差最小 思路: 在原来的数列开头添加一个0 每次找到的区间为 [min(i,j)+1,max(i,j)] 应用尺取法的代码: while (r <=n) { int sub = pre[r].sum - pre[l].sum; while (abs(…

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…

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…

//yy:因为这题多组数据,DP预处理存储状态比每次尺取快多了,但是我更喜欢这个尺取的思想. 题目链接:codeforces 814 C. An impassioned circulation of affection 题意:给出字符串长度n (1 ≤ n ≤ 1 500),字符串s由小写字母组成,q个询问q (1 ≤ q ≤ 200 000),每个询问为:mi (1 ≤ mi ≤ n)   ci ,表示可以把任意mi个字母改成ci,求每次改变后只由ci组成的最长连续字串的长度. 解法一:尺取法…

Subsequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18795 Accepted: 8043 Description A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are gi…