H-Bomb

最新推荐文章于 2023-08-30 14:42:58 发布

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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

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.

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49
int 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4
			sta2 = 1;
		if (sta == 1)//有4
		{
			if (i == 4)//又是4
				sta2 = 1;
			else if (i == 9)//出现49
				sta2 = 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;
}