HDU3555 Bomb —— 数位DP

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+7;
const int maxn = 100+10;

LL n;
LL a[maxn], dp[maxn][3];

LL dfs(int pos, int status, bool lim)
{
    if(!pos) return status==2;
    if(!lim && dp[pos][status]!=-1) return dp[pos][status];

    LL ret = 0;
    int m = lim?a[pos]:9;
    for(int i = 0; i<=m; i++)
    {
        int tmp_status;
        if(status==2 || status==1 && i==9)
            tmp_status = 2;
        else if(i==4)
            tmp_status = 1;
        else
            tmp_status = 0;

        ret += dfs(pos-1, 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 = 0;
        while(n)
        {
            a[++p] = n%10;
            n /= 10;
        }
        memset(dp,-1, sizeof(dp));
        LL ans = dfs(p, 0, 1);
        printf("%lld
",ans);
    }
}