Bomb

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

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.

思路：

dp[i][0]表示不含49的个数 ；

dp[i][1]表示不含49且高位为9的个数；

dp[i][2]表示包含49的个数；

dp[i][0]=10*dp[i-1][0]-dp[i-1][1];  //不含49的数可以任意加10个数字，减去9前面加4的个数

dp[i][1]=dp[i-1][0];                //不含49的数最高位加9

dp[i][2]=dp[i-1][2]*10+dp[i-1][i]; //包含49的数字可以添加0~9，高位为9的可以加4；

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
#include"math.h"
#include"vector"
using namespace std;
#define LL __int64
#define N 25
LL dp[N][3];
void inti()
{
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    int i;
    for(i=1; i<N; 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];
    }
}
LL work(LL x)
{
    int i,k,flag;
    int a[N];
    LL ans;
    k=0;
    while(x)
    {
        a[++k]=x%10;
        x/=10;
    }
    a[k+1]=ans=flag=0;
    for(i=k; i>0; i--)
    {
        ans+=a[i]*dp[i-1][2];
        if(flag)
            ans+=a[i]*dp[i-1][0];
        else
        {
            if(a[i]>4)
                ans+=dp[i-1][1];
        }
        if(a[i+1]==4&&a[i]==9)
            flag=1;
        //     printf("%d\n",ans);
    }
    return ans;
}
int main()
{
    int T;
    LL n;
    inti();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d",&n);
        printf("%I64d\n",work(n+1));
    }
    return 0;
}