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hdu3555 Bomb(数位dp)

时间:2017-05-10 01:05:45      阅读:217      评论:0      收藏:0      [点我收藏+]

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                                                   Bomb

                                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
                                                                                                             Total Submission(s): 18060    Accepted Submission(s): 6639

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
数位dp入门 睡觉  明天解释
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cctype>
 5 #include<cmath>
 6 #include<cstring>
 7 #include<map>
 8 #include<stack>
 9 #include<set>
10 #include<vector>
11 #include<algorithm>
12 #include<string.h>
13 typedef long long ll;
14 typedef unsigned long long LL;
15 using namespace std;
16 const int INF=0x3f3f3f3f;
17 const double eps=0.0000000001;
18 ll N,dp[33][3],sum;
19 int bit[33];
20 int main()
21 {
22     int i,j,T,len;
23     memset(dp,0,sizeof(dp));
24     dp[0][0]=1;
25     for(i=1;i<=20;i++){
26         dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
27         dp[i][1]=dp[i-1][0];
28         dp[i][2]=dp[i-1][1]+dp[i-1][2]*10;
29     }
30     scanf("%d",&T);
31     while(T--){
32         scanf("%I64d",&N);
33         N++;
34         len=sum=0;
35         while(N){
36             bit[++len]=N%10;
37             N/=10;
38         }
39         bit[len+1]=0;
40         bool flag=0;
41         for(i=len;i>=1;i--){
42             sum+=dp[i-1][2]*bit[i];
43             if(!flag&&bit[i]>4)  sum+=dp[i-1][1];
44             if(flag)    sum+=dp[i-1][0]*bit[i];
45             if(bit[i+1]==4&&bit[i]==9)  flag=1;
46         }
47         cout<<sum<<endl;
48     }
49     return 0;
50 }

 

 

hdu3555 Bomb(数位dp)

标签:ace   arch   imp   acm-icpc   amp   sub   icp   gre   ota   

原文地址:http://www.cnblogs.com/Aa1039510121/p/6833480.html

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