Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15372 Accepted Submission(s): 5563

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

数位DP，之前做了一个相反的题目，结果WA了，后来发现数据不对，是long long 型数据。

#include <stdio.h>
#include <string.h>

int bit[25];
long long dp[25][2];

long long dfs(int len,bool is4,bool ismax)
{
    if(len==0) return 1;     ///搜索成功
    if(!ismax&&dp[len][is4]>=0) return dp[len][is4];

    long long cnt = 0;
    int maxnum = ismax? bit[len]:9;
    for(int i=0; i<=maxnum; i++)
    {
        if((is4&&i==9)) continue;
        cnt +=dfs(len-1,i==4,ismax&&i==maxnum);
    }
    return ismax?cnt:dp[len][is4]=cnt;
}

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

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long n;
        scanf("%lld",&n);
        memset(dp,-1,sizeof(dp));
        printf("%lld
",n-f(n)+1);
    }
    return 0;
}