（数学+尺取法）2739 - Sum of Consecutive Prime Numbers

Posted 2020-08-06 tak_fate

原题链接：http://poj.org/problem?id=2739

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题意：问一个数有几种方法用连续的素数和表示。

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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++。

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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 }