hdu 3555 bomb（数位dp）

最新推荐文章于 2021-01-22 20:25:28 发布

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本文介绍了一种算法，用于计算从1到N中包含子序列'49'的数字数量，以此来确定时间炸弹的最终威力点数。通过递归深度优先搜索的方法，实现了高效的求解。

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 20583 Accepted Submission(s): 7705

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

Sample Output

0
1
15

HintFrom 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<algorithm>
#include<string.h>
using namespace std;
__int64 dp[50][2];
__int64 a[50];
__int64 dfs(__int64 pos,__int64 pre,__int64 sta,bool limit)
{
	if(pos==-1)
	{
		return 1;
	}
	if(dp[pos][sta]!=-1&&!limit)
	return dp[pos][sta];
	
	__int64 up=limit?a[pos]:9;
	__int64 i,sum=0;
	for(i=0;i<=up;i++)
	{
		if(pre==4&&i==9)
		continue;
		sum+=dfs(pos-1,i,i==4,limit&&i==up);
	} 
	if(!limit)
	dp[pos][sta]=sum;
	return sum;
}
__int64 solve(__int64 x)
{
	__int64 cnt=0;
	while(x)
	{
		a[cnt++]=x%10;
		x=x/10;
	}
	return dfs(cnt-1,-1,0,1);
}
int main(int argc, char *argv[])
{
	int t;
	scanf("%d",&t);
	memset(dp,-1,sizeof(dp));
	while(t--)
	{
		__int64 x;
		scanf("%I64d",&x);
		printf("%I64d\n",x+1-solve(x));//0-x的数为 x+1  solve（x）为所有统计的非满足 49 的数
	}
	return 0;
}