hdu3555--Bomb（数位dp练习3）

Bomb

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

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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.
         

求1到n中不含49的数的个数

dp[i][j][0]代表i位的数，最高位为j时，没有出现49的个数

dp[i][j][1]代表i位的数，最高位为j时，出现49的个数

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
LL dp[25][10][2] ;
int digit[25] , cnt ;
void init()
{
    int i , j , k ;
    memset(dp,0,sizeof(dp)) ;
    for(j = 0 ; j < 10 ; j++)
        dp[1][j][0] = 1 ;
    for(i = 2 ; i <= 20 ; i++)
    {
        for(j = 0 ; j < 10 ; j++)
        {
            for(k = 0 ; k < 10 ; k++)
            {
                dp[i][j][1] += dp[i-1][k][1] ;
                if( j == 4 && k == 9 )
                    dp[i][j][1] += dp[i-1][k][0] ;
                else
                    dp[i][j][0] += dp[i-1][k][0] ;
            }
            //printf("dp[%d][%d]  %I64d %I64d\n", i , j , dp[i][j][0], dp[i][j][1]) ;
        }
    }
    return ;
}
LL solve(LL temp)
{
    int flag , i , j ;
    LL ans = 0 ;
    memset(digit,0,sizeof(digit)) ;
    cnt = 0 ;
    while( temp )
    {
        digit[++cnt] = temp % 10 ;
        temp /= 10 ;
    }
    for(j = 0 ; j < digit[cnt] ; j++)
        ans += dp[cnt][j][1] ;
    flag = 0 ;
    for(i = cnt-1 ; i > 0 ; i--)
    {
        for(j = 0 ; j < digit[i] ; j++)
        {
            ans += dp[i][j][1] ;
            if( flag )
                ans += dp[i][j][0] ;
        }
        if( digit[i+1] == 4 && digit[i] == 9 )
            flag = 1 ;
    }
    return ans ;
}
int main()
{
    LL t , n , ans ;
    init() ;
    scanf("%I64d", &t) ;
    while( t-- )
    {
        scanf("%I64d", &n) ;
        ans = solve(n+1) ;
        printf("%I64d\n", ans) ;
    }
    return 0;
}