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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define in(n) scanf("%d",&(n))
#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))
#define inll(n) scanf("%I64d",&(n))
#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))
#define inlld(n) scanf("%lld",&(n))
#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))
#define inf(n) scanf("%f",&(n))
#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))
#define inlf(n) scanf("%lf",&(n))
#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))
#define inc(str) scanf("%c",&(str))
#define ins(str) scanf("%s",(str))
#define out(x) printf("%d\n",(x))
#define out2(x1,x2) printf("%d %d\n",(x1),(x2))
#define outf(x) printf("%f\n",(x))
#define outlf(x) printf("%lf\n",(x))
#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));
#define outll(x) printf("%I64d\n",(x))
#define outlld(x) printf("%lld\n",(x))
#define outc(str) printf("%c\n",(str))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mem(X,Y) memset(X,Y,sizeof(X));
typedef vector<int> vec;
typedef long long ll;
typedef pair<int,int> P;
const int dx[]={,,-,},dy[]={,,,-};
const int INF=0x3f3f3f3f;
const ll mod=1e9+;
ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true; bool is_prime[];
int prime[];
void table(){
int k=;
fill(is_prime,is_prime+,true);
rep(i,,){
if(is_prime[i]) prime[k++]=i;
for(int j=*i;j<=;j+=i)
is_prime[j]=false;
}
}
int main(){
int n,s,t,sum,ans;
table();
while(in(n)==){
if(n==) break;
s=t=;ans=;
sum=;
while(true){
while(prime[t]<=n&&sum<n){
sum+=prime[t++];
}
if(sum<n) break;
else if(sum==n) ans++;
sum-=prime[s++];
}
out(ans);
}
return ;
}

POJ2739 Sum of Consecutive Prime Numbers(尺取法)的更多相关文章

  1. poj 2739 Sum of Consecutive Prime Numbers 尺取法

    Time Limit: 1000MS   Memory Limit: 65536K Description Some positive integers can be represented by a ...

  2. POJ2739 Sum of Consecutive Prime Numbers 2017-05-31 09:33 47人阅读 评论(0) 收藏

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25225 ...

  3. POJ2739 Sum of Consecutive Prime Numbers 确定某个数以内的所有素数

    参考:https://www.cnblogs.com/baozou/articles/4481191.html #include <iostream> #include <cstdi ...

  4. POJ2739 - Sum of Consecutive Prime Numbers(素数问题)

    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<cstrin ...

  5. 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 ...

  6. Sum of Consecutive Prime Numbers(poj2739)

    Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22019 Ac ...

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

    Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 27853 Ac ...

  8. poj 2379 Sum of Consecutive Prime Numbers

                                                                                                        ...

  9. POJ 2739. Sum of Consecutive Prime Numbers

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050 ...

随机推荐

  1. JS函数定义与匿名函数的调用

    一.函数声明.函数表达式.匿名函数 函数声明:function fnName () {…};使用function关键字 声明一个函数,再指定一个函数名,叫函数声明. 函数表达式 var fnName ...

  2. linux“批处理”脚本

    依次执行A B C3条语句 最简单的 写如下代码 #!/bin/sh A B C 保存为test.sh然后添加执行权限chmod +x test.sh然后执行该脚本./test.sh

  3. MongoDB资料汇总专题[转发]

    转发下..这个哥收集的很全 MongoDB资料汇总专题 作者:nosqlfan http://blog.nosqlfan.com/html/3548.html 最后更新时间:2013-04-22 1. ...

  4. 清除oracl中有主外键关联的表中的部分数据。

    1.禁用主外键BEGINfor c in (select 'ALTER TABLE '||TABLE_NAME||' DISABLE CONSTRAINT '||constraint_name||' ...

  5. Applet: 用HTML调用Applet的几个注意事项

    问题:HTML找不到java class. 首先,如果xxx.java文件与HTML文件在同一目录下,直接运行cmd-javac 该 xxx.java文件,生成xxx.class文件.HTML中的&l ...

  6. Linux_jdk path (execute and install)

    作者:潇湘隐者 出处:http://www.cnblogs.com/kerrycode/ 1:echo $JAVA_HOME 使用$JAVA_HOME的话能定位JDK的安装路径的前提是配置了环境变量$ ...

  7. 1160 蛇形矩阵(codevs)

    http://codevs.cn/problem/1160/ 题目描述 Description 小明玩一个数字游戏,取个n行n列数字矩阵(其中n为不超过100的奇数),数字的填补方法为:在矩阵中心从1 ...

  8. Android 读取和保存文件(手机内置存储器)

    1:activity_main.xml <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/androi ...

  9. Attach file to database

    D:\Program Files\Microsoft SQL Server\MSSQL11.MSSQLSERVER\MSSQL\DATA databaseName.mdf databaseName.l ...

  10. c#调用c++开发的dll const char* 返回值接收问题

    原文:c#调用c++开发的dll const char* 返回值接收问题 用c#调用视频接口相关的dll,dll使用c++开发. c++接口定义如下: PLATFORM const char* Pla ...