Bomb HDU

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

3150500

Sample Output

0115          

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.

#include <stdio.h>#include <iostream>#include <algorithm>#include <cmath>#include <map>#include <vector>using namespace std;int a[25];long long dp[25][2][12];//当前位数，是否已包含49，前一位是几    //      当前位置     是否包含49。  前一位      限制位long long dfs(int pos, bool have, int pre,  bool limit) {        //如果已枚举完，并且已经包含49 则输出1，否则0了。    if (pos == -1) return have;        //记忆化dp的正确姿势    if (!limit && dp[pos][have][pre] != -1) return dp[pos][have][pre];    int up = limit ? a[pos] : 9;    long long temp = 0;        for (int i = 0; i <= up; ++i) {        if (pre == 4 && i == 9) {//存在49的状态            temp += dfs(pos-1, true, i, limit && i == a[pos]);        } else {                 //不存在49的状态            temp += dfs(pos-1, have, i, limit && i == a[pos]);        }    }        if (!limit)        dp[pos][have][pre] = temp;    return temp;}long long solve(long long x) {    int i = 0;    while (x) {        a[i++] = x%10;        x /= 10;    }    return dfs(i-1, false, 0, 1);}int main() {    int T;    long long shit;    cin >> T;    while (T--) {        memset(dp, -1, sizeof(dp));        cin >>shit;        cout << solve(shit) << endl;    }}