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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- Sum of Consecutive Prime Numbers(poj2739)
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22019 Ac ...
- POJ:2739-Sum of Consecutive Prime Numbers(尺取)
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 27853 Ac ...
- poj 2379 Sum of Consecutive Prime Numbers
...
- 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
- Swift语法之 ---- ?和!区别
1.常量和变量 Swift语言中是用let来定义常量,并且要初始化.var来定义变量,在let或者var后面申明类型,冒号+空格,然后再加上类型名称. 2.optional(可选)变量 可选变量用于处 ...
- android更新SDK时候丢失annotations.jar 导致支持库报错
I am trying to update my Android SDK Tools to 17 rev. and I updated usign SDK Tools but in Propertie ...
- Cracking the coding interview--Q1.5
原文 Implement a method to perform basic string compression using the counts of repeated characters. F ...
- ChangeServiceConfig2 function
ChangeServiceConfig2 function Changes the optional configuration parameters of a service. Syntax C ...
- COJ 0026 汉诺塔问题
汉诺塔问题 难度级别:B: 运行时间限制:1000ms: 运行空间限制:51200KB: 代码长度限制:2000000B 试题描述 经典的汉诺塔游戏相信很多同学都会玩的,规则就不用赘述,百科一下就OK ...
- S-Nim
http://acm.hdu.edu.cn/showproblem.php?pid=1536 SG经典题,不多说 // File Name: hdu1536.cpp // Author: bo_jwo ...
- IntelliJ IDEA创建web项目及异常问题解决
IDEA配置Tomcat: 1.下载Tomcat,本次使用的是apache-tomcat-6.0.43 2.IDEA配置Tomcat 在idea中的Settings(Ctrl+Alt+s)(或者点击图 ...
- RequireJS进阶(二)
这一篇来认识下打包工具的paths参数,在入门一中就介绍了require.config方法的paths参数.用来配置jquery模块的文件名(jQuery作为AMD模块时id为“jquery”,但文件 ...
- 关于Spring配置 (Cannot find class [org.apache.commons.dbcp.BasicDataSource] 问题)
myeclipse抛出异常 org.springframework.beans.factory.CannotLoadBeanClassException: Cannot find /webapps/t ...
- appium 使用findElementByAndroidUIAutomator 定位元素示例
appium 使用findElementByAndroidUIAutomator 定位元素示例 import io.appium.java_client.remote.MobileCapability ...