不一样的二分；
思路：
*二分的上限下限是：MAX（最大的数字），SUM （总和）；
*其中一定有一个数字满足，将N分成M份；
*求最小的；只要当它达到M份时，依然减小上届值；
算法不难，思路想到了就 水到渠成了；
/*
*problem ID: POJ 3273
* Author ID: fuqiang
* TIME: 2013-07-17
* Algorithm: 二分
* Status: (Accept)
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <iomanip>
#define maxn 100010
int a[maxn],N,M;
using namespace std;
inline bool check(int mid)
{
int num = 1, sum = 0;
for(int i = 0; i <= N; i++)
{
if(sum + a[i] <= mid)
sum += a[i];
else
{
sum = a[i];
num++;
}
}
return num>M;
}
int main(int argc, char *argv[])
{
while(scanf("%d%d",&N,&M)!=EOF)
{
int MAX = 0, sum = 0;
for(int i = 1; i <= N; i++)
{
scanf("%d",&a[i]);
sum += a[i];
MAX = max(MAX,a[i]);
}
int left = MAX, right = sum, mid;
while(left < right)
{
mid = (left + right) / 2;
if(check(mid))
left = mid + 1;
else
right = mid - 1;
}
printf("%d\n",left);
}
return 0;
}