数位dp

数位问题：一般有三种状态，各个状态可以用前一个状态表示，此外一般从高位往低位方向处理

HDU  3555:Bomb
Bomb


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 16443    Accepted Submission(s): 6020

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

题意：找出1-N之间包含49的数字的个数

/** Author:xiaomo* Date:2016.12.06* Problem:找出1-n中包含49的个数* Solution:dp[i][0] represent the numbers of not include 49 which length is i;dp[i][1] represent the numbers of not include 49 but start with 9,such as 91241;dp[i][2] represent the numbers of include 49 which length id ithe status change function is:dp[i][0]=10*dp[i-1][0]-dp[i-1][1],if the number is started with 4 and length is idp[i][1]=dp[i-1][0]dp[i][2]=10*dp[i-1][2]+dp[i-1][1]*/#include<iostream>using namespace std;typedef long long lint;const int LENGTH = 21;lint dp[LENGTH][3];void init(){dp[0][0] = 1;for (int i = 1; i < LENGTH; i++){dp[i][0] = 10 * dp[i - 1][0] - dp[i - 1][1];            //不包括49dp[i][1] = dp[i - 1][0];                                //以9开头且不包含49dp[i][2] = 10 * dp[i - 1][2] + dp[i - 1][1];            //包含49}}lint work(lint number){int digit[21] = { 0 };int index = 0;while (number > 0){digit[++index] = number % 10;number /= 10;}digit[index + 1] = 0;lint cnt = 0;bool flag = false;for (int i = index; i > 0; --i)    //from the highest order to lowest order{cnt += digit[i] * dp[i - 1][2];   //from 0-(digit[i]-1),the lower order include 49  if (flag)                      //the higher order include 49{cnt += digit[i] * dp[i - 1][0];}else{if (digit[i] > 4) {cnt += dp[i - 1][1];}}if (digit[i + 1] == 4 && digit[i] == 9){flag = true;}}return cnt;}int main(){int length;cin >> length;lint number;init();for (int i = 0; i < length; i++){cin >> number;cout << work(number + 1) << endl;   //pay attention:because include the number itself}return 0;}