zoukankan      html  css  js  c++  java
  • poj 2739 Sum of Consecutive Prime Numbers(尺取法)

    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

    Source

     
     
    尺取法的两种写法,仅供参考
     
    1、
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 using namespace std;
     5 int prime[100006];
     6 int num[100006];
     7 int sum[100006];
     8 int tot;
     9 void init(){
    10     sum[0]=0;
    11     for(int i=2;i<100006;i++){
    12         if(!num[i]){
    13             prime[tot]=i;
    14             sum[tot]=sum[tot-1]+prime[tot];
    15             tot++;
    16             for(int j=i;j<100006;j+=i){
    17                 num[j]=1;
    18             }
    19         }
    20     }
    21     tot--;
    22 }
    23 int main()
    24 {
    25     tot=1;
    26     init();
    27 
    28     int n;
    29     while(scanf("%d",&n)==1 && n){
    30         int ans=0;
    31         int s=0,t=1;
    32         //int w=0;
    33 
    34         for(;;){
    35             if(t>tot)
    36               break;
    37               
    38             int w=sum[t]-sum[s];
    39             if(w==n)
    40                 ans++;
    41 
    42             if(prime[t]>n)
    43                 break;
    44             if(w<=n)
    45               t++;
    46             if(w>n)
    47               s++;
    48             if(s==t)
    49               t++;
    50         }
    51 
    52         printf("%d
    ",ans);
    53     }
    54     return 0;
    55 }
    View Code

    2、

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 const int MAX = 100005;
     7 int pri[MAX], vis[MAX];
     8 int n, p;
     9 
    10 void ready()
    11 {
    12     p = 0;
    13     memset(vis, 0, sizeof(vis));
    14     for(int i = 2; i < MAX; ++i) {
    15         if(vis[i]) continue;
    16         pri[p++] = i;
    17         for(int j = i*2; j < MAX; j += i) vis[j] = 1;
    18     }
    19 }
    20 void solve()
    21 {
    22     int sum = 0, k = 0, t = 0;
    23     int ans = 0;
    24     while(1) {
    25         while(t <= p && sum < n) sum += pri[t++];
    26         if(pri[t-1] > n) break;
    27         if(sum == n) ans++;
    28         sum -= pri[k++];
    29     }
    30     printf("%d
    ", ans);
    31 }
    32 int main()
    33 {
    34 //freopen("in", "r", stdin);
    35     ready();
    36     while(~scanf("%d", &n)) {
    37         if(n == 0) break;
    38         solve();
    39     }
    40     return 0;
    41 }
    View Code
     
     
  • 相关阅读:
    c#中的对象生命周期
    数据抓取的艺术(三):抓取Google数据之心得
    redmine3.3.0安装问题
    wget 无法建立ssl连接 [ERROR: certificate common name ?..ssl.fastly.net?.doesn?. match requested host name ?.ache.ruby-lang.org?. To connect to cache.ruby-lang.org insecurely, use ?.-no-check-certificate?]
    Centos安装ruby
    Redmine插件的安装与卸载,知识库插件安装。
    Nexus网页直接上传jar包
    mvn deploy命令上传包
    一辈子只有1次成为BAT的机会,你如何把握?
    redmine创建新闻,自动发邮件给项目组所有成员
  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4767128.html
Copyright © 2011-2022 走看看