Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 7707 | Accepted: 3180 |
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
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day
Output
Sample Input
7 5
100 400 300 100 500 101 400
Sample Output
500
題目大意:知道農民接下來N天每天花的錢,不改變順序要分成M份。求最大一份的最小值是多少。
二分查找,代碼:
#include<iostream> using namespace std; int i,j,n,m,cxb[100010]; bool fine(int a) { int temp=0,x=1; for(i=1;i<=n;i++) { if(x>m) return false; temp+=cxb[i]; if(temp>a) { temp=cxb[i]; x++; } } if(x<=m) return true; } int main() { while(scanf("%d%d",&n,&m)!=EOF) { int sum=0,min=0,max,mid; for(i=1;i<=n;i++) { scanf("%d",&cxb[i]); sum+=cxb[i]; if(cxb[i]>min) min=cxb[i]; //最小爲花的最多的那天 } max=sum;//最大爲總的錢數 int ans; while(max>=min)//進行二分查找 { mid=(max+min)/2; if(fine(mid)) { max=mid-1; ans=mid; } else { min=mid+1; } } printf("%d\n",ans); } return 0; }