hdu3555(数位DP dfs/递推)

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 10649 Accepted Submission(s): 3758

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

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.

Author

fatboy_cw@WHU

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

Recommend

zhouzeyong

采用dfs记忆化搜索暴力枚举，dp[len][0]代表长度为len，不含49，首位可以是9的数的个数，dp[len][1]代表长度为len，不含49,首位不是9的数的数的个数，

这样每次枚举当前位置，

a:如果当前位置是4,那么下一位不能是9，所以dfs(len,1,false/true);

b:如果当前位置不是4，那么下一位没有限制，所以dfs(len,0,false/true);

c:如果包含当前位置的前缀与n这个数相同位置的前缀相等的话，那么下一位的枚举就有限制，所以dfs(len,0/1,true);

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL  long long
LL dp[30][2];
int digit[30];

LL dfs(int len,bool state,bool fp)
{
    if(!len)
        return 1;
    if(!fp && dp[len][state] != -1)
        return dp[len][state];
    LL ret = 0 ;int  fpmax = fp ? digit[len] : 9;
    for(int i=0;i<=fpmax;i++)
    {
        if(state && i == 9)
            continue;
        ret += dfs(len-1,i == 4,fp && i == fpmax);  
        //第二个参数表明这位是否为4，如果为4，下一位就不能为9，否则没有限制
        //第三个参数表明前缀是否相同，如果相同，下一位就只能枚举到最大值，否则就没有限制
    }
    if(!fp)
        dp[len][state] = ret;
    return ret; //ret代表0到n不含49的个数
}

LL f(LL n)
{
    int len = 0;
    while(n)
    {
        digit[++len] = n % 10;
        n /= 10;
    }
    return dfs(len,false,true);
}

int main()
{
    //freopen("test.txt","r",stdin);
    LL a,b;
    memset(dp,-1,sizeof(dp));
    int t;
    scanf("%d",&t);
    while(t--)
    {
          cin>>b;
          cout<<(b+1-f(b))<<endl;
    }

    return 0;
}