Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6609 Accepted Submission(s): 2303
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
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.
第一个数位动规！
！！！
看代码理解的。差点儿相同懂那么点点点了。
AC代码例如以下：
传统模板！
！
！
///递推 46MS 328K
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
__int64 dp[20][3];//0表示该位数中不含49的情况数。1表示不含49但头为9的情况数。2表示含49的情况数
int i,j;
memset(dp,0,sizeof dp);
dp[0][0]=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][0];
dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
//cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl;
}
int t;
cin>>t;
__int64 num[30];
__int64 n;
while(t--)
{
memset(num,0,sizeof num);
cin>>n;
n++;//这一步非常重要
int tt=1;
while(n)
{
num[tt++]=n%10;
n/=10;
}
__int64 ans=0;
int flag=0;int last=0;
for(i=tt-1;i>=1;i--)
{
ans+=num[i]*dp[i-1][2];
if(flag==1)
ans+=dp[i-1][0]*num[i];
if(flag==0&&num[i]>4)
ans+=dp[i-1][1];
if(num[i+1]==4&&num[i]==9)
flag=1;
}
cout<<ans<<endl;
}
return 0;
}
///数位之记忆化搜索 78MS 276K
#include<cstdio>
#include<iostream>
#include<cstring>
#define mod 1000000009
#define ll __int64
#define M 100005
using namespace std;
ll dp[30][3];
ll num[30];
ll dfs(int pos,int statu,int limit)
{
int i;
if(!pos) return statu==2;
if(!limit&&dp[pos][statu]!=0) return dp[pos][statu];
int end=limit?num[pos]:9;
ll sum=0;
for(i=0;i<=end;i++)
{
int flag=statu;
if(flag==0&&i==4) flag=1;
if(flag==1&&i==9) flag=2;
if(flag==1&&i!=4&&i!=9) flag=0;
sum+=dfs(pos-1,flag,limit&&i==end);
}
return limit?
sum:(dp[pos][statu]=sum);
}
ll _49(ll n)
{
int pos=1;
memset(dp,0,sizeof dp);
while (n>0)
{
num[pos++]=n%10;
n/=10;
}
//cout<<pos<<"~~~~~~~~~~~~~"<<endl;
return dfs(pos-1,0,1);
}
int main()
{
int i,j;
int t;
scanf("%d",&t);
while(t--)
{
ll n;
scanf("%I64d",&n);
printf("%I64d
",_49(n));
}
return 0;
}