poj 3273 Monthly Expense(贪心+二分)

题目：http://poj.org/problem?id=3273

题意：把n个数分成m份，使每份的和尽量小，输出最大的那一个的和。

思路：二分枚举最大的和，时间复杂度为O(nlog(sum-max));

一道很好的题。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn = 100000+10;
int a[maxn];

int main()
{
    int n, m, i, Max, sum;
    while(~scanf("%d%d", &n, &m))
    {
        Max = -1;
        sum = 0;
        for(i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
            if(Max < a[i])
            Max = a[i];
        }
        int high = sum, low = Max, mid;
        while(high>low)
        {
            mid = (high+low)/2;
            int cou = 1, w = 0;
            for(i = 0; i < n; i++)
            {
                w += a[i];
                if(w>mid)
                {
                    cou++;  //cou记录最大值为mid的情况下，可以分成几份
                    w = a[i];
                }
            }
            if(cou>m)
            low = mid+1;
            else
            high = mid-1;
        }
        printf("%d\n", high);
    }
    return 0;
}

 

 

以后还是用这种形式的二分吧

while(left<right)

{

if()

left=mid+1;

else right=mid;

}

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn = 100000+10;
int a[maxn];

int main()
{
    int n, m, i, Max, sum;
    while(~scanf("%d%d", &n, &m))
    {
        Max = -1;
        sum = 0;
        for(i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
            if(Max < a[i])
            Max = a[i];
        }
        int high = sum, low = Max, mid;
        while(high > low)
        {
            mid = (high+low)/2;
            int cou = 1, w = 0;
            for(i = 0; i < n; i++)
            {
                w += a[i];
                if(w>mid)
                {
                    cou++;
                    w = a[i];
                }
            }
            if(cou <= m)
            high = mid;
            else
            low = mid+1;
        }
        printf("%d\n", low);
    }
    return 0;
}