H-Bomb

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.        

      T组，每次给一个数n，让求1-n之间有多少个数中含有49。

      还是一个模板套下来，不过值得一提的是，这个是第一个完完全全脱了稿做出来的数位dp题目，哈哈，虽然是这个题简单，但是原来这么轻松，真是提自信了。不过没有一A还是很遗憾，没有注意到n的值是2^64-1，第一次用了LL确忘了把n定义成LL，而却dp[]、a[]都开小了点，换大了就好了。

代码如下：

#include <iostream>#include <cstring>using namespace std;typedef long long LL;LL dp[100][3];//sta   0: 末尾不是4   1：末尾是4   2: 有49int a[100];LL ans;LL dfs(int pos, int sta, bool limit){if (pos == -1){if (sta == 2)return 1;return 0;}if (!limit && dp[pos][sta] != -1)return dp[pos][sta];int up = limit ? a[pos] : 9;LL temp = 0;for (int i = 0; i <= up; i++){int sta2 = sta;if (sta == 0 && i == 4)//不是4出现4sta2 = 1;if (sta == 1)//有4{if (i == 4)//又是4sta2 = 1;else if (i == 9)//出现49sta2 = 2;else//不是49/44，出现4*sta2 = 0;}temp += dfs(pos - 1, sta2, limit && i == a[pos]);}if (!limit)dp[pos][sta] = temp;return temp;}LL solve(LL x){int pos = 0;while (x){a[pos++] = x % 10;x /= 10;}return dfs(pos - 1, 0, true);}int main(){int T;LL n;memset(dp, -1, sizeof(dp));cin >> T;while (T--){cin >> n;memset(a, 0, sizeof(a));LL ans = solve(n);cout << ans << endl;}return 0;}