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.
Lines 2.. N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdio> 6 #include<algorithm> 7 #define inf 9999999999 8 #define LL long long 9 using namespace std; 10 int a[100011]; 11 int low,high,mid; 12 int n,m,i,j,k; 13 int main() 14 { 15 scanf("%d%d",&n,&m); 16 high=0; 17 low=-inf; 18 for(i=1;i<=n;i++) 19 { 20 scanf("%d",&a[i]); 21 high+=a[i]; 22 if(low<a[i]) 23 low=a[i]; 24 } 25 int get_m,sum; 26 while(low<high) 27 { 28 mid=(low+high)/2; 29 sum=0; 30 get_m=0; 31 for(i=1;i<=n;i++) 32 { 33 if(sum+a[i]<=mid) 34 { 35 sum+=a[i]; 36 } 37 else 38 { 39 sum=a[i]; 40 get_m++; 41 } 42 } 43 if(get_m<m)high=mid; 44 else 45 low=mid+1; 46 } 47 printf("%d\n",low); 48 }
Reproduced in: https: //www.cnblogs.com/sdau--codeants/p/3380676.html