（数位dp）Bomb （hdu 3555）

http://acm.hdu.edu.cn/showproblem.php?pid=3555

 

 

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.

 

Author

fatboy_cw@WHU

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

 

 

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;

LL dp[30][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9
 * dp[i][2],表示长度为i, 存在不吉利数字
 */

 void Init()
 {
     int i;

     dp[0][0] = 1;
     for(i=1; i<30; i++)
     {
         dp[i][0] = 10*dp[i-1][0] - dp[i-1][1];
         dp[i][1] = dp[i-1][0];
         dp[i][2] = 10*dp[i-1][2] + dp[i-1][1];
     }
 }

 LL Slove(LL n)
 {
     int i, a[65]={0}, len=0, flag=0;
     LL n1 = n, ans=0;

     while(n1)
     {
         a[++len] = n1%10;
         n1 /= 10;
     }

     for(i=len; i>=1; i--)
     {
         ans += dp[i-1][2]*a[i];
         if(flag)///高位已经出现49了，后面随意
            ans += dp[i-1][0]*a[i];
         else if(!flag && a[i]>4)
            ans += dp[i-1][1];
         if(a[i+1]==4 && a[i]==9)
            flag = 1;
     }

     return ans;
 }
int main()
{
    int T;
    Init();
    scanf("%d", &T);
    while(T--)
    {
        LL n;

        scanf("%I64d", &n);

        printf("%I64d
", Slove(n+1));
    }
    return 0;
}