POJ_3273(二分)

Description

Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ money_i ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.

FJ wants to create a budget for a sequential set of exactly M (1 ≤ M ≤ N) fiscal periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.

FJ's goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

Input

 

Line 1: Two space-separated integers: N and M
Lines 2.. N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day

 

Output

 

Line 1: The smallest possible monthly limit Farmer John can afford to live with.

 

Sample Input

7 5
100
400
300
100
500
101
400

Sample Output

500

Hint

If Farmer John schedules the months so that the first two days are a month, the third and fourth are a month, and the last three are their own months, he spends at most $500 in any month. Any other method of scheduling gives a larger minimum monthly limit.

 

https://vjudge.net/contest/176507#problem/F

 

题意为，将给出的连续序列分成M份，所有的区间都要小一个值，求出这个值的最小值。

写了四次wa了无数发 终于过了，教训就是多写多用大脑跑程序。

 

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1e5+5;
int a[maxn],n,m,sum,maxa,ans;

int f(int x){
	int sum = 0,cnt = 0;
	for(int i=0;i<n;i++){
		sum+=a[i];
		if(sum>x){
			sum=a[i];
			cnt++;
		}
	}
　　　　　return cnt+1;//防止最后一个区间没有被加上个。
}

int _s(int x,int y){
	int l = x, r = y, mid  = 0;
	while(l<=r){
		mid=(l+r)>>1;
		if(f(mid)<=m) r=mid-1;//注意相等的时候不一定找到最小值，必须要一直分下去，直到r==l；
		else l=mid+1;
	}
	return mid;
}

int main(){
	while(scanf("%d %d",&n,&m)==2){
		sum = maxa  = ans = 0;
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
			maxa=max(maxa,a[i]);
			sum+=a[i];
		}
		ans = _s(maxa,sum);
		printf("%d
",ans);
	}
}