HDU 3555 数位DP

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6471    Accepted Submission(s): 2250


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.

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <string.h>
 4 #include <iostream>
 5 using namespace std;
 6 
 7 main()
 8 {
 9     __int64 n, dp[33][4], ans;
10     int t, i, j, k, bit[33];
11     memset(dp,0,sizeof(dp));
12     dp[0][0]=1;
13     for(i=1;i<=20;i++)                             //dp过程 
14     {
15         dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
16         dp[i][1]=dp[i-1][0];
17         dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
18     }
19     scanf("%d",&t);
20     while(t--)
21     {
22         scanf("%I64d",&n);
23         k=0;
24         n++;                             //因为该程序求的是[0,n)半开半闭区间的个数，以防数字n满足条件，所以n+1 
25         while(n)                                 //把n每位数字放进一个数组里 
26         {
27             bit[++k]=n%10;
28             n/=10;
29         }
30         ans=0;
31         bit[k+1]=0;
32         int flag=0;
33         for(i=k;i>=1;i--)                      //求解过程 
34         {
35             ans+=dp[i-1][2]*bit[i];
36             if(flag) ans+=dp[i-1][0]*bit[i];            //如果高位数字中含有49 那么以后不管是什么数字都满足条件 
37             if(!flag&&bit[i]>4) ans+=dp[i-1][1];         //如果高位数字钟不含49 但是第i位大于4，那么需要加上第i位是4，第i+1位是9的数字 
38             if(bit[i+1]==4&&bit[i]==9) flag=1;
39         }
40         printf("%I64d\n",ans);
41     }
42 }