ural Threeprime Numbers(dp)

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http://acm.timus.ru/problem.aspx?space=1&num=1586

题意没看懂，看了别人的翻译。threeprime number的意思是任意三个连续的数组成的一个三位数是素数，注意必须是三位数。给出n，问满足条件的n位数有多少个。

先把三位数的素数筛选出来并标记，设dp[i][j][k]表示到i位为止，最后两位是j和k的满足条件的数的个数。

那么dp[i][j][k] += dp[i-1][g][j]。

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL __int64#define eps 1e-12#define PI acos(-1.0)#define PP pair<LL,LL>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 1000;const int mod = 1000000009;int prime[maxn+10];int flag[maxn+10];int cnt = 0;int tprime[maxn+10];int dp[10010][12][12];void get_prime(){memset(flag,0,sizeof(flag));prime[0] = 0;flag[0] = flag[1] = 1;for(int i = 2; i <= maxn; i++){if(flag[i] == 0){prime[++prime[0]] = i;if(i >= 100 && i <= 999)tprime[++cnt] = i;}for(int j = 1; j <= prime[0] && i*prime[j] <= maxn; j++){flag[prime[j]*i] = 1;if(i % prime[j] == 0)break;}}}int main(){get_prime();int n;while(~scanf("%d",&n)){if(n == 3){printf("%d\n",cnt);continue;}memset(dp,0,sizeof(dp));for(int i = 1; i <= cnt; i++){int b = tprime[i]%10;int a = tprime[i]/10%10;dp[3][a][b] += 1;}for(int i = 4; i <= n; i++){for(int j = 0; j <= 9; j++){for(int k = 0; k <= 9; k++){for(int g = 0; g <= 9; g++){int num = j*100 + k*10 + g;if(flag[num] == 0 && num >= 100) //num必须是三位数{dp[i][k][g] += dp[i-1][j][k]%mod;dp[i][k][g] %= mod;}}}}}LL ans = 0;for(int i = 0; i <= 9; i++){for(int j = 0; j <= 9; j++)ans = (ans + dp[n][i][j])%mod;}printf("%I64d\n",ans);}return 0;}

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