Bomb

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.

 

 

 

开始看不懂题的时候可以看一下后面的举例子，做题的时候没有看到例子好久才看懂的，说多了都是泪。

 

题目大意：

　　给一个数字n,范围在1~2^63-1，求1~n之间含有49的数字有多少个。

经典的数位dp，参考了大腿的代码：http://www.cnblogs.com/luyi0619/archive/2011/04/29/2033117.html

 

状态转移：

　　dp[i][0]代表长度为 i 并且不含有49的数字的个数；

 

　　dp[i][1]代表长度为 i 并且不含有49，但是最高位是9的数字的个数；

 

　　dp[i][2]代表长度为 i 并且含有49的数字的个数。

 

　　数组 a[i] 从低位到高位存储 n 的每一位数字。

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

 

代码如下：

#include<iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll dp[25][3];


void init()
{
    dp[0][0] = 1; //表示位数为i的无49的个数; 等于1是为首位为9，位数为1的数服务；
    dp[0][1] = 0; //表示位数为i的无49的个数但最高位为9的个数；
    dp[0][2] = 0; //表示位数为i的含有49的个数；
    for(int i = 1; i < 25; i++)
    {
        dp[i][0] = dp[i - 1][0] * 10 - dp[i - 1][1]; //减去最高位为9的情况；
        dp[i][1] = dp[i - 1][0];                     //最高位加9的个数
        dp[i][2] = dp[i - 1][2] * 10 + dp[i - 1][1]; //加上没含49但最高位是9的个数（加上4就成49了）；
    }
}
ll solve(ll n)
{
    ll ans = 0;
    bool vis = false;
    int bit[25],tot = 0;
    while(n)
    {
        bit[++tot] = n % 10;
        n /= 10;
    }
bit[tot + 1] = 0;
    for(int i = tot; i >= 1; i--)
    {
        ans += dp[i - 1][2] * bit[i];
        if(vis) ans += bit[i] * dp[i - 1][0]; // 这一步和下一步要细心体会其中的高深之处，看了kuangbin大神的代码好久才明白这个道理；
        else if(bit[i] > 4) ans += dp[i - 1][1]; //如果n本身已经含有49了就不用考虑是否最高位为9了，直接加上没49的个数，因为n含了49
        if(bit[i + 1] == 4 && bit[i] == 9) vis = true;
    }
    if(vis) ans++; //如果n本身也包含49，则要加1；
    return ans;
}


int main()
{
    ll n;
    int t;
    init();
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld",&n);
        printf("%lld\n",solve(n));
    }
    return 0;
}