（数学+尺取法）2739 - Sum of Consecutive Prime Numbers
原题链接：http://poj.org/problem?id=2739
题意：问一个数有几种方法用连续的素数和表示。
分析：其实就是很简单，先打表，然后对prime数组跑一波尺取法，如果==n就ans++。
代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cmath> 5 #include <cstring> 6 #include <set> 7 #include <vector> 8 #include <queue> 9 #include <map> 10 #include <list> 11 #include <bitset> 12 #include <string> 13 #include <cctype> 14 #include <cstdlib> 15 16 using namespace std; 17 18 typedef long long ll; 19 typedef unsigned long long ull; 20 21 22 const int inf = 1<<30; 23 const ll lnf = 1ll<<60; 24 25 //-------------------------- 26 27 const int maxn=10010; 28 int prime[maxn]; 29 bool vis[maxn]; 30 31 int Euler_prime() { 32 memset(vis, true, sizeof(vis)); 33 int tot = 0; 34 for (int i = 2; i < maxn; i++) { 35 if (vis[i]) prime[tot++] = i; 36 for (int j = 0; j < tot&&prime[j] * i < maxn; j++) { 37 vis[i*prime[j]] = false; 38 if (i%prime[j] == 0) break; 39 } 40 } 41 return tot; 42 } 43 44 45 void solve() { 46 int n; 47 int pn=Euler_prime(); 48 while(~scanf("%d",&n)&&n){ 49 int ans=0; 50 int l=0,r=0; 51 int res=0; 52 while(r<pn){ 53 while(res<n){ 54 res+=prime[r]; 55 r++; 56 if(r>=pn)break; 57 } 58 while(res>=n){ 59 if(res==n)ans++; 60 res-=prime[l]; 61 l++; 62 } 63 } 64 printf("%d\n",ans); 65 } 66 } 67 68 69 70 int main() { 71 72 #ifndef ONLINE_JUDGE 73 freopen("in.txt", "r", stdin); 74 //freopen("out.txt", "w", stdout); 75 #endif 76 //iostream::sync_with_stdio(false); 77 solve(); 78 return 0; 79 }
