HDU 3555 Bomb （数位DP）

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.

求含49的个数：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
//ll dp[22][0]; //含有49
//ll dp[22][1]; //不含49且最高位为9
//ll dp[22][2]; //不含49
ll dp[22][3];
int bits[22];

void init()
{
int i,j;
dp[0][2]=1;
for(i=1;i<=20;i++) {
dp[i][0]=dp[i-1][0]*10+dp[i-1][1];
dp[i][1]=dp[i-1][2];
dp[i][2]=dp[i-1][2]*10-dp[i-1][1];
}
}

ll work(ll n)
{
int i,j,len=0;
bool flag=false;
ll ans=0;
while(n) {
bits[++len]=n%10;
n=n/10;
}
bits[len+1]=0;
for(i=len;i;i--) {
ans+=bits[i]*dp[i-1][0];
if(flag) ans+=bits[i]*dp[i-1][2];
if(!flag&&bits[i]>4) ans+=dp[i-1][1];
if(bits[i]==9&&bits[i+1]==4) flag=true;
}
return ans;
}

int main()
{
int t,i,j;
ll n;
cin>>t;
init();
while(t--) {
cin>>n;
cout<<work(n+1)<<endl;
}
return 0;
}