题目大意是求给出的N个数分成连续的M组， 求组内数之和的最小值。
发现二分循环里最后输出mid比较保险。。一开始输出了l总是错
题目：
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 11653 | Accepted: 4769 |
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
Hint
Source
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 #define MAXN 100000+100 5 #define min(x,y) x<y?x:y 6 int N,M; 7 int cost[MAXN]; 8 9 bool C(int x) 10 { 11 int n = 1; 12 // cout<<x<<endl; 13 int tmp =0; 14 for(int s = 0; s<N;s++) 15 { 16 if( tmp+cost[s]<=x) 17 { 18 tmp+=cost[s]; 19 } 20 else 21 { 22 n++; 23 tmp = cost[s]; 24 } 25 } 26 if( n>M)return false; 27 28 else return true; 29 } 30 31 32 int main() 33 { 34 while(cin>>N>>M) 35 { 36 int l = 0 , r=0; 37 for(int i=0;i<N;i++) 38 { 39 cin>>cost[i]; 40 r+= cost[i]; 41 l = max(cost[i],l); 42 } 43 int mid = l+(r-l)/2; 44 while( l<r) 45 { 46 47 //cout<<l<<" "<<r<<endl; 48 if( C(mid) ) r= mid-1; 49 else 50 l = mid+1; 51 mid = l+(r-l)/2; 52 //cout<<mid<<endl; 53 } 54 cout<<mid<<endl; 55 } 56 return 0; 57 }