西交校赛 I. GZP and CS(数位dp)

I. GZP and CS

GZP love to play Counter-Strike(CS).

One day GZP was playing a mod of CS.

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 "96", the power of the blast would add one point.

Now GZP knows the number N. Only when the computing result of the final points of the power by GZP is correct can he attain success.Can you help him?

 

Input

The first line of input consists of an integer T (1 <= T <= 100), indicating the number of test cases.  For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

 

Output

For each test case, output an integer indicating the final points of the power.

 

Sample Input

3
1
100
500

Sample Output

0
1
5

Hint

From 1 to 500, the numbers that include the sub-sequence "96" are "96","196","296","396","496",so the answer is 5.

 

题意：

问[1,n]中有多少个数中包含96;

 

思路：

数位dp的题,跟有一道hdu的数位dp一样;哎,还没理解记忆化搜索的写法;有时间看看去;

 

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e6+5e5;
LL dp[25][3],n;
int b[25];
//dp[i][0]表示长度<=i包含96的个数;
//dp[i][1]表示长度为i不包含96但开头为6的个数
//dp[i][2]表示<=i不包含96的个数;
int fun()
{
    mst(dp,0);
    dp[0][2]=1LL;
    for(int i=1;i<23;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];
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    fun();
    while(t--)
    {
        scanf("%lld",&n);
        LL temp=n,ans=0;
        int cnt=1;
        while(temp)
        {
            b[cnt++]=temp%10;
            temp/=10;
        }
        b[cnt]=0;
        int flag=0;
        for(int i=cnt;i>0;i--)
        {
            ans+=dp[i-1][0]*(LL)b[i];
            if(flag)//如果前边已经出现了96那么就还要加上后面不含96的方案数;
            {
                ans+=dp[i-1][2]*b[i];
            }
            if(b[i+1]==9&&b[i]==6)
            {
                flag=1;
            }
        }
        if(flag)ans++;//如果96出现的位置后面全是0
        printf("%lld\n",ans);
    }
    return 0;
}