O-Bomb(数位dp)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15062    Accepted Submission(s): 5437

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 t 组数据，在 1- n 中出现了49的数据个数，n 很大，2^63-1 , 遍历肯定不行。

 

//听说这是数位dp的一道水题，这也是水题吗，我服了

自己想了一阵，只想到了一点点。看了别人的也看了很久，才明白，这篇博客非常详细，我就不赘述了，认真看一定能看懂。

http://www.cnblogs.com/liuxueyang/archive/2013/04/14/3020032.html

我的与他的dp数组有细微差别，因为我自己理解了写了一遍，稍微注意一下。

 

代码 15ms

 

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

__int64 dp[25][3];
//dp[i][0]   含49，数的个数
//dp[i][1] 不含49，但首位是9的数的个数
//dp[i][2] 不含49，数的个数(包含了首位是9的数的个数)
void Init()
{
    dp[0][2]=1;
    for (int i=1;i<=24;i++)
    {
        dp[i][0]=dp[i-1][0]*10+dp[i-1][1];
        dp[i][1]=dp[i-1][2];
        dp[i][2]=dp[i-1][2]*10-dp[i-1][1];
    }
}

__int64 solve(__int64 n)
{
    __int64 a[25],len=0;

    while (n)
    {
        a[++len]=n%10;
        n/=10;
    }
    a[len+1]=0;

    __int64 ans=0;
    int flag=0;

    for (int i=len;i>0;i--)
    {
        ans+=dp[i-1][0]*a[i];
        if (flag)           //前面有49就所有情况都是49数了
            ans+=dp[i-1][2]*a[i];
        if (!flag&&a[i]>4)  //加上4  9... 的情况
            ans+=dp[i-1][1];
        if (a[i]==9&&a[i+1]==4)//判断是不是 49 
            flag=1;
    }
    return ans;
}

int main()
{
    int t;
    __int64 n;
    Init();
    scanf("%d",&t);
    while (t--)
    {
        scanf("%I64d",&n);
        printf("%I64d
",solve(n+1));//因为函数功能是 (0,n) 区间的49数，题目要求的是[1,n]，所以+1
    }
    return 0;
}