HDU 3555 Bomb (数位dp)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 18031    Accepted Submission(s): 6632


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

HintFrom 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

 

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题目的意思是给你一个数，判断从1到这个数之间有子串49的数有多少个.

数据比较大，需要long long.

数位dp的入门题目

#include <iostream>
#include<string>
#include<cstring>
#include<set>
#include<cstdio>
using namespace std;
typedef long long ll;

ll dp[25][3];
ll a[25];

ll dfs(int pos,bool is,int state,bool limit)
{
if(pos==-1&&state==1) return 1;
else if(pos==-1) return 0;
if(!limit&&dp[pos][state]!=-1) return dp[pos][state];
int up=limit?a[pos]:9;
ll temp=0;
for(int i=0;i<=up;i++)
{
if((is&&i==9)||state==1)
temp+=dfs(pos-1,i==4,1,limit&&a[pos]==i);
else if(i==4) temp+=dfs(pos-1,i==4,2,limit&&a[pos]==i);
else temp+=dfs(pos-1,i==4,0,limit&&a[pos]==i);
}
if(!limit) dp[pos][state]=temp;
return temp;
}

ll solve(ll x)
{
memset(dp,-1,sizeof(dp));
int pos=0;
while(x)
{

a[pos++]=x%10;
x/=10;
}
return dfs(pos-1,0,0,true);
}

int main()
{
ll x,y;
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
scanf("%lld",&x);
cout<<solve(x)<<endl;
}
return 0;
}