【HDU 3486】【二分+RMQ】 Interviewe 【最小化区间个数m，使得每个区间最大值和大于k】

传送门：HDU 3486 Interviewe

描述：

Interviewe

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6666    Accepted Submission(s): 1580

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 so busy that he has no time to interview them by himself. So he decides to select exact m interviewers for this task.
YaoYao decides to make the interview as follows. First he queues the interviewees according to their coming order. Then he cuts the queue into m segments. The length of each segment is  , 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.
YaoYao’s idea seems to be wonderful, but he meets another problem. He values the ability of the ith arrived interviewee as a number from 0 to 1000. Of course, the better one is, the higher ability value one has. He wants his employees good enough, so the sum of the ability values of his employees must exceed his target k (exceed means strictly large than). On the other hand, he wants to employ as less people as possible because of the high salary nowadays. Could you help him to find the smallest m?

 

Input

The input consists of multiple cases.
In the first line of each case, there are two numbers n and k, indicating the number of the original people and the sum of the ability values of employees YaoYao wants to hire (n≤200000, k≤1000000000). In the second line, there are n numbers v1, v2, …, vn (each number is between 0 and 1000), indicating the ability value of each arrived interviewee respectively.
The input ends up with two negative numbers, which should not be processed as a case.

 

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.

 

Sample Input

  

   11 300
7 100 7 101 100 100 9 100 100 110 110
-1 -1
  

 

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.
   
    
  

 

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

 

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题意：

给定n个数的序列，让我们找前面m个区间的最大值之和，每个区间长度为n/m，如果有剩余的区间长度不足n/m则无视之。

现在让我们找最小的m使得和严格大于k。n<=20万，k <= 10亿。

思路：

用二分枚举k，然后用RMQ来查询区间内的最大值，即可

代码：

#include <bits/stdc++.h>
#define pr(x) cout << #x << "= " << x << "  " ；
#define pl(x) cout << #x << "= " << x << endl;
#define ll __int64
using  namespace  std;

#define mst(ss,b) memset(ss,b,sizeof(ss));
#define For(i,j,n) for(int i=j;i<=n;i++)

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
const int N=2e5+10;

int n,k;
int a[N];
int dp[N][20];

int rmq_init(){
  for(int i=1; i<=n; i++)dp[i][0]=a[i];
  for(int j=1; (1<<j)<=n; j++)
    for(int i=1; i+(1<<j)-1<=n; i++)
      dp[i][j]=max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
}

int rmq(int l,int r){
  int k=(int)(log(r-l+1.0)/log(2.0));
  return max(dp[l][k], dp[r-(1<<k)+1][k]);
}

int cal(int x, int y){
  int res=0;
  for(int i=1; i<=y; i++){
    res+=rmq((i-1)*x+1, i*x);
    if(res>k)return res;
  }
  return res;
}

int  main(){
  #ifndef ONLINE_JUDGE
  freopen("in.txt","r",stdin);
  #endif

  while(~scanf("%d%d",&n,&k)){
    if(n<0 && k<0)break;
    int sum=0,mx=0;
    for(int i=1; i<=n; i++)read(a[i]),sum+=a[i],mx=max(mx, a[i]);
    if(mx>k) { puts("1");continue;}//先预判断一下区间为1和n的情况
    if(sum<=k) { puts("-1"); continue;}
    mst(dp, 0);
    rmq_init();
    int l=1,r=n,mid,t,ans; 
    while(l<=r){//二分区间个数
      mid=(l+r)>>1;
      t=cal(n/mid, mid);//区间长度,几个区间
      if(t>k)r=mid-1,ans=mid;
      else l=mid+1;
    }   
    printf("%d\n",ans);
  }
  return 0;
}