HDU 3555 数位dp

时间：2023-03-08 16:16:05

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

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 ;

 }

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