zoj 1539 Lot 简单DP 记忆化

最新推荐文章于 2021-07-22 10:52:20 发布

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本文介绍了一种算法，用于从一排士兵中选择不同组合的三名侦察兵。通过迭代移除奇数或偶数位置上的士兵，直到剩下三人。文章提供了完整的C++代码实现，并说明了输入输出的要求。

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Lot

Time Limit: 2 Seconds Memory Limit: 65536 KB

Out of N soldiers, standing in one line, it is required to choose several to send them scouting.
In order to do that, the following operation is performed several times: if the line consists of more than three soldiers, then all soldiers, standing on even positions, or all soldiers, standing on odd positions, are taken away. The above is done until three or less soldiers are left in the line. They are sent scouting. Find, how many different groups of three scouts may be created this way.

Note: Groups with less than three number of soldiers are not taken into consideration.

0 < N <= 10 000 000

Input

The input file contains the number N.

Process to the end of file.

Output

The output file must contain the solution - the amount of variants.

Sample Input

10
4

Sample Output

2
0

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int f[10000001];
int _find(int v)
{
	if(v<3) return f[v]=0;
	if(v==3) return f[v]=1;
	if(f[v]>0) return f[v];
	if(v%2==0)  return f[v]=_find(v/2)*2;
	else  return f[v]=_find(v/2)+_find(v/2+1);
}
int main()
{
	int n,s;
	while(cin>>n)//不知道为什么，这里换成输入优化就超超时了。。。。。。求大佬告知
	{
		s=_find(n);
		if(s>0)
		printf("%d\n",s);
		else printf("0\n");
	}
	return 0;
}