[原创]HDU 3555 Bomb [数位DP]【动态规划】

2016-08-22 20:34:43 Tabris_ 阅读数：366

博客爬取于2020-06-14 22:43:45
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52280031

题目链接 ：HDU最近比较炸 不贴链接了、、
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Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15374 Accepted Submission(s): 5565

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.

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题目大意 ： 就是问你从1~m 里 有49的数的个数

解题思路 :

最最基础的数位DP

附本题代码
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# include <bits/stdc++.h>

using namespace std;

# define LL long long int

LL  dp[30][12][2];
//    位数  数字0~9  状态
int num[30],len ;
//  数字   长度





/**********记忆化搜索****************/
//分别代表现在的位数 上一位的数字 限制  状态

/*
主要就是限制
比如说m = 123456789;
那么位数到地5位的时候 就不能选6 7 ..等等了
但是如果在第5位他选择了 4  那么下一位就能够任意选择了 、
*/
/*
状态就是表示 到现在 有没有他能够满足的状态  有就+1  没有就+0;ll
*/

LL dfs(int pos ,int x,int limit ,int status)
{
    if(pos<0)  return status;

    if(dp[pos][x][status]!=-1&&!limit)
        return dp[pos][x][status];

    int endi=9;
    if(limit) endi = num[pos];

    LL res = 0;
    for(int i=0;i<=endi;i++)
    {
        res+=dfs(pos-1 ,i ,limit&&(i==endi) ,status||(i==9&&x==4) );
    }

    if(!limit) dp[pos][x][status] =  res;
    return  res;
}


int main()
{

    memset(dp,-1,sizeof(dp));
    int _;
    scanf("%d",&_);
    while(_--)
    {
        LL n;
        scanf("%I64d",&n);
        LL tem = n;
        len = 0;
        while(tem)
        {
            num[len++]=tem%10;
            tem/=10;
        }
        LL m = dfs(len-1,0,1,0);

        printf("%I64d\n",m);

    }
    return 0;
}