【HDU 3555】Bomb 数位dp模板

Bomb

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)[/b]
[b]Total Submission(s): 16611    Accepted Submission(s): 6082[/b]

[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.

模板即可。

输出输出、dp数组和ans记得开long long。

#include
#include
#include
#include
#include
#include
using namespace std;
long long dp[18][3];
int a[10];
long long dfs(int now,int before,int doing)
{
int i;
if(!doing&&dp[now][before]!=-1)
return dp[now][before];
if(now==0)
return 1;
long long ans=0;
int s=doing?a[now]:9;
for(i=0;i<=s;i++)
{
if(i==9&&before)continue;
ans+=dfs(now-1,i==4,doing&&i==s);
}
if(!doing)dp[now][before]=ans;
return ans;
}
long long solve(long long n)
{
int len=0;
memset(a,0,sizeof(a));
do
{
a[++len]=n%10;
n/=10;
}while(n);
return dfs(len,0,1);
}
int main()
{
int T;
long long n;
scanf("%d",&T);
while(T--)
{
scanf("%lld",&n);
memset(dp,-1,sizeof(dp));
printf("%lld\n",(n+1-solve(n)));
}
}