HDU 3555 Bomb 数位dp

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3716 Accepted Submission(s): 1298


[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

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.
代码加注释：

//数位DP//包含多少个49
#include<iostream>
#include<cstdio>
using namespace std ;
long long dp[35][3] ;
int len[25] ;
void init()
{
dp[0][0]=1;
dp[0][1]=dp[0][2]=0;
for(int i=1;i<25;i++)
{
dp[i][0]=10*dp[i-1][0]-dp[i-1][1];//在前面加0~9的数字，减掉在9前面加4
dp[i][1]=dp[i-1][0];//最高位加9
dp[i][2]=10*dp[i-1][2]+dp[i-1][1];//在本来含有49的前面加任意数，或者在9前面加4
}
}
long long dfs( long long n)
{
int i , s = 0 , j , ok = 0 ;
long long ans = 0 ;
while(n)// 把各个位存到数组里面
{
len[++s] = n % 10 ;
n /= 10 ;
}
len[s+1] = 0 ;
for( i = s ; i >= 1 ;i-- )
{
ans += dp[i-1][2] * len[i] ; // 下一个是49所以这位都是符合答案的
if(ok) ans += dp[i-1][0] * len[i] ;//
else
{   // 如果是大于4 则 需 下位是9
if(len[i]>4) ans+=dp[i-1][1];
}
// 上位是4 ，这位是9 ,下次来都是
if(len[i+1]==4&&len[i]==9)ok = 1 ;
}
if(ok)ans++ ; //最后加上最后一位
return ans ;
}
int main()
{
init() ;
int i , T ;
long long n ;
cin >> T ;
while( T-- )
{
cin >> n ;
cout << dfs(n) << endl ;
}
}