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; }