POJ3273 Monthly Expense

Monthly Expense Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25959   Accepted: 10021 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

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题目的意思是给出n个数，分成5组，要求每组之和最大值最小是多少

思路：二分+验证

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long

LL a[100005];
int n,m;
bool ok(LL x)
{
    LL sum=0;
    int cnt=0;
    for(int i=0; i<n; i++)
    {
        sum+=a[i];
        if(sum>x)
        {
            cnt++;
            sum=a[i];
        }
    }
    if(sum>0)
        cnt++;
    if(cnt<=m)
        return 1;
    return 0;
}


int main()
{

    while(~scanf("%d%d",&n,&m))
    {
        LL mx=0;
        LL sum=0;
        for(int i=0; i<n; i++)
        {
            scanf("%lld",&a[i]);
            mx=max(mx,a[i]);
            sum+=a[i];
        }

        LL l=mx,r=sum;
        LL ans;
        while(l<=r)
        {
            LL mid=(l+r)/2;
            if(ok(mid))
            {
                r=mid-1;
                ans=mid;
            }
            else
            {
                l=mid+1;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}