NOJ——1627Alex’s Game(II)（尺取）

• [1627] Alex’s Game(II)

• 时间限制: 2000 ms　内存限制: 65535 K
• 问题描述
• Alex likes to play with one and zero as you know .

  Today he gets a sequence which contains n(n<=1e5) integers.Each integer is no nore than 100.now he wants to know what’s the minimun contigous subsequence that their puduct contain no less than k(k<=1e5) zeros in the tail.

• 输入
• First are two integers n and k.
  Next contains n lines ,every line contains n integers. 
  All integers are bigger than 0.
• 输出
• For each case output the answer if you can find it.
  Else just output “haha”
• 样例输入

• 5 1
  2 10 2 5 1
  5 3
  2 10 2 5 1

• 样例输出

• 1
  haha

刚开始以为K=0的话就直接输出0，后来发现0的情况要特判：若每一个数都末尾带0，即能被10整除，则输出0，否则只要取那个不是10倍数的数即可，输出1.其他情况用尺取法。

统计2与5的个数，只有这两个数可以影响末尾0的个数，每一次加上L或减掉R那边的两个统计个数即可。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
int two[100010];
int five[100010];
int pos[100010];
inline int conttwo(int n)
{
	if(n%2!=0)
		return 0;
	else
	{
		int ans=0;
		while (n%2==0)
		{
			ans++;
			n>>=1;
		}
		return ans;
	}
}
inline int contfive(int n)
{
	if(n%5!=0)
		return 0;
	else
	{
		int ans=0;
		while (n%5==0)
		{
			ans++;
			n/=5;
		}
		return ans;
	}
}
int main(void)
{
	int n,k,dx,ans,i,j,t;
	while (~scanf("%d%d",&n,&k))
	{
		memset(two,0,sizeof(two));
		memset(five,0,sizeof(five));
		for (i=1; i<=n; i++)
		{
			scanf("%d",&pos[i]);
			two[i]=conttwo(pos[i]);
			five[i]=contfive(pos[i]);
		}
		if(k<=0)//特判0
		{
			bool flag=0;
			for (i=1; i<=n; i++)
			{
				if(pos[i]%10!=0)
				{
					flag=1;
					break;
				}
			}
			if(flag)
				printf("%d\n",1);
			else
				printf("%d\n",0);
			continue;
		}
		int l=1,r=1,dx=INF;
		int sum5=0,sum2=0;
		while (1)
		{
			while (r<=n&&min(sum5,sum2)<k)
			{
				sum5+=five[r];
				sum2+=two[r];
				r++;
			}
			if(min(sum5,sum2)<k)
				break;
			dx=min(dx,r-l);
			sum5-=five[l];
			sum2-=two[l];
			l++;
		}
		if(dx==INF)
			printf("haha\n");
		else
			printf("%d\n",dx);
	}
	return 0;
}