URAL 1586

题目大意：求出n位十进制数中每相邻三位均为一个三位数素数的个数对10^9+9取模的值。

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

数据规模：3<=n<=10000。

理论基础：无。

题目分析：1万位，不会是高精度吧？Come on,孩子,别忘了这道题的分类是dp。好了，言归正传。

首先，我们用dp[i][j][k]表示i位十进制数中满足要求的且最高位为j,次高位为k的数的个数。bool数组pre[i][j][k]表示三位分别为ijk时是不是素数。

然后，初始化，很显然，dp[3][j][k]=sum(pre[j][k][m],1<=m<=9)。

下来，我们可以很轻易的得出状态转移方程：dp[i][j][k]=sum(pre[j][k][m]*dp[i-1][k][m],1<=m<=9)对10^9+9取模。

最终的答案即为:ans=sum(dp[n][j][k],0<=k<=9,0<j<=9)对10^9+9取模。

代码如下：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<ctime>
#include<vector>
using namespace std;
typedef double db;
#define DBG 0
#define maa (1<<31)
#define mii ((1<<31)-1)
#define ast(b) if(DBG && !(b)) { printf("%d!!|
", __LINE__); while(1) getchar(); }  //调试
#define dout DBG && cout << __LINE__ << ">>| "
#define pr(x) #x"=" << (x) << " | "
#define mk(x) DBG && cout << __LINE__ << "**| "#x << endl
#define pra(arr, a, b)  if(DBG) {
    dout<<#arr"[] |" <<endl; 
    for(int i=a,i_b=b;i<=i_b;i++) cout<<"["<<i<<"]="<<arr[i]<<" |"<<((i-(a)+1)%8?" ":"
"); 
    if((b-a+1)%8) puts("");
}
template<class T> inline bool updateMin(T& a, T b) { return a>b? a=b, true: false; }
template<class T> inline bool updateMax(T& a, T b) { return a<b? a=b, true: false; }
typedef long long LL;
typedef long unsigned int LU;
typedef long long unsigned int LLU;
#define N 10000
#define M 10
int dp[N+1][M][M],n;
bool pre[M][M][M]={
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,
0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,
0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,
0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,
0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,
0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,
0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,
0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,
0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,
0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,
0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,
0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,
0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,
0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,
0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,
0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,
0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,
0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,
0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,
0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,
0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,
0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,
0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,
0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,
0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,
0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,
0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,
0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,
0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,
0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0};
int nprime[143]={101,
103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,
211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,
331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,
449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,
587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,
709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,
853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,
991,997};
int main()
{
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof dp);
        for(int i=0;i<=142;i++)
        {
            dp[3][nprime[i]/100][nprime[i]/10%10]++;
        }
        for(int i=4;i<=n;i++)
        {
            for(int j=1;j<=9;j++)
            {
                for(int k=0;k<=9;k++)
                {
                    for(int l=1;l<=9;l+=2)
                    {
                        if(pre[j][k][l])dp[i][j][k]=(dp[i-1][k][l]+dp[i][j][k])%1000000009;
                    }
                }
            }
        }
        int ans=0;
        for(int j=1;j<=9;j++)
        {
            for(int k=0;k<=9;k++)
            {
                ans=(ans+dp[n][j][k])%1000000009;
            }
        }
        printf("%d
",ans);
    }
	return 0;
}

其中，pre[i][j][k]的初始化可以依靠nprime[]数组得到。

by:Jsun_moon http://blog.csdn.net/jsun_moon