HODJ3555 Bomb【数位dp】

Bomb

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

Total Submission(s): 10459    Accepted Submission(s): 3684


[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?
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.题意求1-n中包含49的个数

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int digital[25];
long long dp[25][3];
void dpdabiao()
{
int i,j;
dp[0][0]=1;
for(i=1;i<=23;++i)
{
dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//不含49的个数
dp[i][1]=dp[i-1][0];//最高位为9的个数
dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//含有49的个数
}
}
int main()
{
dpdabiao();
int i,j,t;
long long n;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
int l=1;
memset(digital,0,sizeof(digital));
while(n)
{
digital[l++]=n%10;
n/=10;
}
bool sign=false;
long long count=0;
for(i=l-1;i>=1;--i)
{
count+=dp[i-1][2]*digital[i];
if(sign)
{
count+=dp[i-1][0]*digital[i];
}
else
{
if(digital[i]>4)
count+=dp[i-1][1];
if(digital[i+1]==4&&digital[i]==9)sign=true;

}
}if(sign)count++;
printf("%lld\n",count);
}
return 0;
}