数位DP的DFS写法。。。。
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4630 Accepted Submission(s): 1614
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
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
Recommend
zhouzeyong
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long int LL;
int bit[30];
LL dp[25][3];
LL dfs(int pos,int s,bool limit)
{
if(pos==-1)
return s==2;
if(limit==false&&~dp[pos][s])
return dp[pos][s];
int end=limit?bit[pos]:9;
LL ans=0;
for(int i=0;i<=end;i++)
{
int news=s;
if(s==0&&i==4) news=1;
if(s==1&&i!=9) news=0;
if(s==1&&i==9) news=2;
if(s==1&&i==4) news=1;
ans+=dfs(pos-1,news,limit&&i==end);
}
if(!limit)
return dp[pos][s]=ans;
else
return ans;
}
int main()
{
LL n;
int T;
memset(dp,-1,sizeof(dp));
scanf("%d",&T);
while(T--)
{
scanf("%I64d",&n);
int len=0;
while(n)
{
bit[len++]=n%10;
n/=10;
}
printf("%I64d\n",dfs(len-1,0,1));
}
return 0;
}
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