HDU-3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 11029    Accepted Submission(s): 3916

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.

 

/**
    题意：1~n之间有多少数包含49
    做法：数位dp
    ///dp[i][0] 表示当前位没有49
    ///dp[i][1] 表示当前位是以9开头
    ///dp[i][2] 表示含有49
    设sum = 0；
    对于n进行从高到低进行的扫描
    对于当前位 sum += dp[i-1][2]*mmap[i] 对于i-1满足要求的
    如果当前的位数 > 4 并且没有包含49 那么 sum += dp[i-1][1];
    如果当前已经有49了 那么 sum += dp[i-1][0] * mmap[i];
**/
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long dp[][];
int digit[];
///dp[i][0] 表示当前位没有49
///dp[i][1] 表示当前位是以9开头
///dp[i][2] 表示含有49
int main()
{
    memset(dp, , sizeof(dp));
    dp[][] = ;
    for(int i = ; i < ; i++) ///预处理  很好理解
    {
        dp[i][] = dp[i - ][] *  - dp[i - ][];
        dp[i][] = dp[i - ][];
        dp[i][] = dp[i - ][] *  + dp[i - ][];
    }
    int t;
    cin >> t;
    while(t--)
    {
        int len = , last = ;
        long long ans = ;
        unsigned long long n = ;
        cin >> n;
        n++;
        memset(digit, , sizeof(digit));
        while(n)
        {   digit[++len] = n % ;
            n /= ;
        }
        bool flag = false;
        for(int i = len; i >= ; i--)
        {
            ans += dp[i - ][] * digit[i]; ///当前位的可能
            if(flag)
            {
                ans += dp[i - ][] * digit[i]; ///如果已经含有49 加上I-1个不符合要求的个数
            }
            if(!flag && digit[i] > )   //当前位 > 4  以为前一位已经是9了所以包含49
            {
                ans += dp[i - ][];
            }
            if(last ==  && digit[i] == ) ///包含49
            {   flag = true;
            }
            last = digit[i];    ///标记上一位
        }
        cout << ans << endl;
    }
    return ;
}