Monthly Expense Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 38688 Accepted: 14256
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 ≤ moneyi ≤ 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. Source
USACO 2007 March Silver
二分，需要动动脑子，写法不唯一，二分不能学太死。
代码语言：javascript复制#include <cstdio>
#include <iostream>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi= acos(-1.0);
const int maxn=100010;
int n,m;
int a[maxn];
int judge(int value)
{
int i;
int sum=0,cnt=1;
for(i=0;i<n;i ){
if(sum a[i]<=value)
sum=sum a[i];
else{
sum=a[i];
cnt ;
}
}
if(cnt>m)
return 0;
else
return 1;
}
int main()
{
int high,low,mid;
while(~scanf("%d %d",&n,&m)){
low=high=0;
for(int i=0;i<n;i ){
scanf("%d",&a[i]);
low=max(low,a[i]);
high =a[i];
}
while(low<=high){
mid=(low high)>>1;
if(judge(mid))
high=mid-1;
else
low=mid 1;
}
printf("%d",mid);
}
return 0;
}
