hdu 3555 Bomb

Bomb

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

Total Submission(s): 6563    Accepted Submission(s): 2288


[align=left]Problem Description[/align]
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?

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, output an integer indicating the final points of the power.
 

[align=left]Sample Input[/align]

3
1
50
500

 

[align=left]Sample Output[/align]

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.

求1~n包含49的数的个数，简单数位dp

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define LL long long
int p[25];
LL dp[25][2][10];
LL DP(int h,bool ok,int last,bool sign){
if(h<0) return ok;
if(!sign && dp[h][ok][last]!=-1) return dp[h][ok][last];
LL ans = 0;
if(sign){
for(int i=0;i<=p[h];i++){
if(last == 4 && i==9) ans += DP(h-1,true,i,i==p[h]);
else ans += DP(h-1,ok,i,i==p[h]);
}
}
else{
for(int i=0;i<=9;i++){
if(last==4 && i==9) ans += DP(h-1,true,i,false);
else ans += DP(h-1,ok,i,false);
}
}
if(!sign) dp[h][ok][last]=ans;
return ans;
}
int main(){
LL T,n;
cin >> T;

while(T--){
memset(dp,-1,sizeof dp);
cin >> n;
int cnt=0,t=n;
while(n>0){
p[cnt++] = n%10;
n /= 10;
}
cout << DP(cnt-1,false,0,true) <<endl;
}
return 0;
}