POJ 3273 Monthly Expense

题意：经典的最大值最小问题……给FJ在n天里每天花的钱数，问如果将n天分成m组，每组内的天数连续，最少每组给多少钱可以让每组的钱够花。

解法：二分。现在看来是个挺简单的二分……然而我二分一向都是理论AC……写出来跟屎一样……QAQ……贴个代码当模板好了……

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include<iomanip>
#define LL long long
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;

int a[100005];
int n, m;
bool judge(int mid)
{
    int cnt = 1;
    int sum = 0;
    for(int i = 0; i < n; i++)
    {
        if(a[i] > mid) return false;
        if(sum + a[i] > mid)
        {
            cnt++;
            sum = a[i];
        }
        else
        {
            sum += a[i];
        }
    }
    return cnt <= m;
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        int l = 0, r = 1000000000;
        int mid = (l + r) >> 1;
        while(l < r)
        {
            mid = (l + r) >> 1;
            if(judge(mid)) r = mid;
            else l = mid + 1;
        }
        cout << l << endl;
    }
    return 0;
}