POJ-2739Sum of Consecutive Prime Numbers（尺取法+埃拉托斯特尼筛法）

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首先了解埃拉托斯特尼筛法

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看完图解基本就知道埃拉托斯特尼筛法的步骤了

知道了打表然后就是水题了

#include <iostream>#include <map>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <queue>#include <stack>#include <functional>#include <set>#include<sstream>#include <cmath>using namespace std;#define pb push_back#define PB pop_back#define bk back()#define fs first#define se second#define INF 1e10;#define sq(x) (x)*(x)#define eps (1e-10)#define clr(x) memset((x),0,sizeof (x))#define cp(a,b) memcpy((a),(b),sizeof (b))typedef long long ll;typedef unsigned long long ull;typedef pair<int,int> P;const int maxn=10100;int is_prim[maxn];int prim[maxn];int k=1;void get_prim(){    int m=sqrt(maxn)+1;    for(int i=2;i<maxn;i++) is_prim[i]=1;    for(int i=2;i<m;i++)    {        if(is_prim[i])        {            for(int j=i+i;j<maxn;j+=i) is_prim[j]=0;        }    }}int main(){    clr(prim);    prim[k++]=2;    get_prim();    for(int i=3;i<maxn;i++) if(is_prim[i]) prim[k++]=i;    k--;    int n;    while(cin>>n&&n)    {        int sum=0,len=0,s=1,f=1,cnt=0;        for(;;)        {            while(sum<n&&s<=k)            {               sum+=prim[s++];               len++;            }            if(len==0) break;            if(sum==n) cnt++;            sum-=prim[f++];            len--;        }        cout<<cnt<<endl;    }    return 0;}

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