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 representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime 
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. 
Your mission is to write a program that reports the number of representations for the given positive integer.

Input

The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10 000, inclusive. The end of the input is indicated by a zero.

Output

The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.

Sample Input

2
3
17
41
20
666
12
53
0

Sample Output

1
1
2
3
0
0
1
2
思路:尺取法操作连续子序列。
import java.util.Arrays;
import java.util.Scanner; public class Main {
Scanner in = new Scanner(System.in);
final int MAXN = 10005;
int[] prime = new int[MAXN];
boolean[] isPrime = new boolean[MAXN];
int[] sum = new int[MAXN];
int total;
void table() {
Arrays.fill(isPrime, true);
isPrime[0] = false;
isPrime[1] = false;
for(int i = 2; i < MAXN; i++) {
if(isPrime[i]) {
prime[total++] = i;
for(int j = i + i; j < MAXN; j += i) {
isPrime[j] = false;
}
}
}
sum[0] = 0;
for(int i = 1; i < total; i++) {
sum[i] = sum[i-1] + prime[i-1];
}
}
Main() {
int n;
table();
while((n = in.nextInt()) != 0) {
int res = 0, sum = 0;
int front = 0, rear = 0;
while(true) {
while(rear < total && prime[rear] <= n && sum < n) {
sum += prime[rear++];
}
if(sum == n) {
res++;
}
sum -= prime[front++];
if(front >= total || front > rear) {
break;
}
}
System.out.println(res);
}
}
public static void main(String[] args) { new Main();
}
}

POJ2739(尺取法)的更多相关文章

  1. poj2739尺取法+素数筛

    Some positive integers can be represented by a sum of one or more consecutive prime numbers. How man ...

  2. POJ2739 Sum of Consecutive Prime Numbers(尺取法)

    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 ...

  3. poj2739(尺取法+质数筛)

    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<algor ...

  4. POJ 尺取法

    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, ...

  5. 5806 NanoApe Loves Sequence Ⅱ(尺取法)

    传送门 NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K ...

  6. POJ3061 尺取法

    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 ...

  7. POJ 2739 Sum of Consecutive Prime Numbers(尺取法)

    题目链接: 传送门 Sum of Consecutive Prime Numbers Time Limit: 1000MS     Memory Limit: 65536K Description S ...

  8. CF 701C They Are Everywhere(尺取法)

    题目链接: 传送门 They Are Everywhere time limit per test:2 second     memory limit per test:256 megabytes D ...

  9. nyoj133_子序列_离散化_尺取法

    子序列 时间限制:3000 ms  |  内存限制:65535 KB 难度:5   描述 给定一个序列,请你求出该序列的一个连续的子序列,使原串中出现的所有元素皆在该子序列中出现过至少1次. 如2 8 ...

随机推荐

  1. linux中的/usr,/var,/opt目录详解

    转自:http://it.greenblogs.org/archives/2008/20113.shtml/ /usr文件系统  /usr 文件系统经常很大,因为所有程序安装在这里. /usr 里的所 ...

  2. 【网络编程】inet_addr、inet_ntoa、inet_aton、inet_ntop和inet_pton区分

    先上一张图 1.把ip地址转化为用于网络传输的二进制数值 int inet_aton(const char *cp, struct in_addr *inp); inet_aton() 转换网络主机地 ...

  3. CF 916

    题解: 首先看题目 A题看不懂... 花了5分钟才做出来 还wa了 B题 一看好像是堆+位运算? 然后A了样例 C题 wa了好激发 似乎加边加错了 然后看D,似乎是可持久化平衡树? 我又不会... E ...

  4. yii2 实现excel导出功能

    官方教程地址:http://www.yiiframework.com/extension/yii2-export2excel/ 安装: Either run php composer.phar req ...

  5. web运行异常解决

    端口占用: 在dos下,输入  netstat   -ano|findstr  8080 //说明:查看占用8080端口的进程 显示占用端口的进程 taskkill  /pid  6856  /f   ...

  6. DRF中序列化器定义及使用

    首先需要明白序列化和反序列化的定义及作用: 序列化是将程序语言转换为JSON/XML; 反序列化是将JSON/XML转换为程序语言; 对应到Django中,序列化即把模型对象转换为字典形式, 在返回给 ...

  7. UNIX发展史(BSD,GNU,linux)(转)

    转自 UNIX发展史(BSD,GNU,linux) 这篇文章写的非常好,在这里转一下. 先前的一個理想 UNIX 系统自 1969 年 Ken ThompsonKen Thompson 与 Denni ...

  8. ffmpeg新老接口对比

    http://blog.csdn.net/leixiaohua1020/article/details/41013567

  9. d3.js(v5.7)树状图

    一.新建画布 二.数据处理 三.绘制连接线 图示: 四.绘制节点.文字 图示: 五.总结 path元素:其实就是定义了绘图的坐标点,从哪开始,移动到哪,怎样移动(命令) 具体可百度(或许以后我会总结一 ...

  10. ubuntu16扩展屏设置

    new ubuntu system setting - Expansion screen settings. 1,System Settings–>Displays 1,set big scre ...