hdu3555Bomb【数位dp入门题】

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 12286    Accepted Submission(s): 4410

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

3150500

 

Sample Output

0115
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

发现dp题多数是签到题啊==

貌似数位dp暂时发现了两种题型，本质还是一种没错==之前的n进制是需要把原数字转化成n叉数，截取自己需要的部分。

这次说的题是针对这个数字本身来说的，也可以认为是一个十进制，就是一个10叉树，我怎么这么机智==用递推解释就是把一个树杈接到当前节点后面形成新的树杈。以这个题为例，dp[i][0]   dp[i][1]   dp[i][2]分别表示i长度下：不存在不吉利数字；不存在不吉利数字开头是9；存在不吉利数字。首先预处理三个数组，然后就简单啦

主要还是不要62那个题没学明白，用的不是标准的数位dp导致的

/*************hdu35552016.3.1115MS1736K1525 BC++**************/#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;long long dp[22][3];/* * dp[i][0],表示长度为i，不存在不吉利数字 * dp[i][1],表示长度为i,不存在不吉利数字，且最高位为2 * dp[i][2],表示长度为i,存在不吉利数字 */void init(){    dp[0][0]=1;dp[0][1]=0;dp[0][2]=0;    for(int i=1;i<=20;i++)    {        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//在最高位加上除4以外的9个数字，但要减掉2之前加上6        dp[i][1]=dp[i-1][0];//在不含不吉利数字的最高位加上2        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];        //在已有不吉利数字前加任意数字，或者无不吉利数字的最高位加4，或者在2前面加6    }}int bit[22];long long solve(long long n){    int len=0;    long long tmp=n;    while(n)    {        bit[++len]=n%10;        n/=10;    }    bit[len+1]=0;    long long ans=0;    bool flag=false;    for(int i=len;i>=1;i--)    {        ans+=dp[i-1][2]*bit[i];        if(flag)//高位已经出现49，后面随意            ans+=dp[i-1][0]*bit[i];        if(!flag&&bit[i]>4)            ans+=dp[i-1][1];        if(bit[i+1]==4&&bit[i]==9)            flag=true;    }    if(flag)ans++;//这个数本身    return ans;}int main(){    //freopen("cin.txt","r",stdin);    //freopen("out.txt","w",stdout);    init();    long long n;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        printf("%I64d\n",solve(n));    }    return 0;}