HDU 3555 数位dp

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15072    Accepted Submission(s): 5441

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

 

题意：求1~n闭区间内含有“49”的数的个数

题解：

dp[i][2] 长度为i 含有“49”的个数

dp[i][1] 长度为i  不含有“49”但是高位为“9”的个数

dp[i][0] 长度为i  不含有“49”的个数

数组 a[i] 从低位到高位存储 n 的每一位数字。

dp[i][2]=dp[i-1][2]*10+dp[i-1][1]; //考虑第i位为“4” i-1位为“9”

dp[i][1]=dp[i-1][0];

dp[i][0]=dp[i-1][0]*10-dp[i-1][1];

对于n处理之前为什么要自增1

因为题目要求处理的是闭区间 可能自增1当作开区间处理

http://www.cnblogs.com/liuxueyang/archive/2013/04/14/3020032.html

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<map>
 #include<algorithm>
 #include<queue>
 #define ll __int64
 using namespace std;
 int t;
 ll n;
 ll a[];
 ll dp[][];
 void init()
 {
     dp[][]=;
     for(int i=; i<=; i++)
     {
         dp[i][]=*dp[i-][]-dp[i-][];
         dp[i][]=dp[i-][];
         dp[i][]=*dp[i-][]+dp[i-][];
     }
 }
 int main()
 {
     init();
     while(scanf("%d",&t)!=EOF)
     {
         for(int i=; i<=t; i++)
         {
             scanf("%I64d",&n);
             memset(a,,sizeof(a));
             int len=;
             n++;
             while(n)
             {
                 a[len]=n%;
                 n=n/;
                 len++;
             }
             int flag=;
             int last=;
             ll ans=;
             for(int j=len; j>=; j--)
             {
                 ans+=dp[j-][]*a[j];
                 if(flag)
                     ans+=dp[j-][]*a[j];
                 if(!flag&&a[j]>)
                     ans+=dp[j-][];
                 if(last==&&a[j]==)
                     flag=;
                 last=a[j];
             }
             printf("%I64d\n",ans);
         }
     }
     return ;
 }