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.
5 1 2 10 2 5 1 5 3 2 10 2 5 1
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;
}
原文：http://www.cnblogs.com/Blackops/p/5413878.html