Interviewe |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 945 Accepted Submission(s): 234 |
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Problem Description YaoYao has a company and he wants to employ m people recently. Since his company is so famous, there are n people coming for the interview. However, YaoYao is |
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Input The input consists of multiple cases. |
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Output For each test case, print only one number indicating the smallest m you can find. If you can’t find any, output -1 instead. |
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Sample Input 11 300 7 100 7 101 100 100 9 100 100 110 110 -1 -1 |
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Sample Output 3 Hint We need 3 interviewers to help YaoYao. The first one interviews people from 1 to 3, the second interviews people from 4 to 6, and the third interviews people from 7 to 9. And the people left will be ignored. And the total value you can get is 100+101+100=301>300. |
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Source 2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU |
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Recommend zhengfeng |
, which means he ignores the rest interviewees (poor guys because they comes late). Then, each segment is assigned to an interviewer and the interviewer chooses the best one from them as the employee./*
一道一道有毒的题，忘了一种竟然可以n个线段
*/
#include<bits/stdc++.h>
using namespace std;
int d[200100][20];
int n,k,cur,ob,maxn;
void init()
{
for (int j = 1; (1 << j) < n; j++){
int t = (1 << j) - 1;
for (int i = 0; i+t < n; i++){
d[i][j] = max(d[i][j-1], d[i+(1<<(j-1))][j-1]);
}
}
}
inline int RMQ(int a, int b)
{
int l = int(log(double(b-a+1))/log(2.0));
return max(d[a][l], d[b+1-(1<<l)][l]);
}
int main()
{
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
while(scanf("%d%d",&n,&k)&&n>=0&&k>=0)
{
ob=0;
maxn=0;
for(int i=0;i<n;i++)
{
scanf("%d",&d[i][0]);
ob+=d[i][0];
maxn=max(maxn,d[i][0]);
}
if(ob<=k)
{
puts("-1");
continue;
}
init();//RMQ预处理
/*二分找最小值*/
int ans=n;
for(int m=max(1,k/maxn);m<n;m++)
{
cur=0;
int t=n/m;//分组的长度
//cout<<"t="<<t<<endl;
for(int i=1;i<=m;i++)//枚举的组数
{
cur+=RMQ(t*(i-1),t*i-1);
//cout<<t*(i-1)+1<<" "<<t*i<<endl;
if(cur>k)
break;
}
if(cur>k)
{
ans=m;
break;
}
}
printf("%d\n",ans);
}
return 0;
}