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  • 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
    ",(x))
    #define out2(x1,x2) printf("%d %d
    ",(x1),(x2))
    #define outf(x) printf("%f
    ",(x))
    #define outlf(x) printf("%lf
    ",(x))
    #define outlf2(x1,x2) printf("%lf %lf
    ",(x1),(x2));
    #define outll(x) printf("%I64d
    ",(x))
    #define outlld(x) printf("%lld
    ",(x))
    #define outc(str) printf("%c
    ",(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[4]={1,0,-1,0},dy[4]={0,1,0,-1};
    const int INF=0x3f3f3f3f;
    const ll mod=1e9+7;
    ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    const bool AC=true;
    
    bool is_prime[10000];
    int prime[5000];
    void table(){
        int k=0;
        fill(is_prime,is_prime+10000,true);
        rep(i,2,10000){
            if(is_prime[i]) prime[k++]=i;
            for(int j=2*i;j<=10000;j+=i)
                is_prime[j]=false;
        }
    }
    int main(){
        int n,s,t,sum,ans;
        table();
        while(in(n)==1){
            if(n==0) break;
            s=t=0;ans=0;
            sum=0;
            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 0;
    }
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  • 原文地址:https://www.cnblogs.com/akrusher/p/5352885.html
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