HDU3555 Bomb —— 数位DP

时间：2022-05-29 03:47:18

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 18739    Accepted Submission(s): 6929

Problem Description
The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would
add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
 
Input
The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

 
Output
For each test case, output an integer indicating the final points of the power.
 
Sample Input
3
1
50
500
 
Sample Output
0
1
15
Hint

From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.

题解：

数位DP通用：dp[pos][sta1][sta2][……]

表示：当前位为pos，之前的状态为sta1*sta2*……stan。n为限制条件的个数。

回到此题，限制条件有两个： 1.上一位是否为4； 2.之前是否已经出现49。

类似题目：http://blog.csdn.net/dolfamingo/article/details/72848001

代码如下：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <cmath>

 #include <queue>

 #include <stack>

 #include <map>

 #include <string>

 #include <set>

 using namespace std;

 typedef long long LL;

 const int INF = 2e9;

 const LL LNF = 9e18;

 const int mod = 1e9+;

 const int maxn = +;

 LL n;

 LL a[maxn], dp[maxn][];

 LL dfs(int pos, int status, bool lim)

 {

     if(!pos) return status==;

     if(!lim && dp[pos][status]!=-) return dp[pos][status];

     LL ret = ;

     int m = lim?a[pos]:;

     for(int i = ; i<=m; i++)

     {

         int tmp_status;

         if(status== || status== && i==)

             tmp_status = ;

         else if(i==)

             tmp_status = ;

         else

             tmp_status = ;

         ret += dfs(pos-, tmp_status, lim&&(i==m));

     }

     if(!lim) dp[pos][status] = ret;

     return ret;

 }

 int main()

 {

     int T;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%lld",&n);

         int p = ;

         while(n)

         {

             a[++p] = n%;

             n /= ;

         }

         memset(dp,-, sizeof(dp));

         LL ans = dfs(p, , );

         printf("%lld\n",ans);

     }

 }

相关文章