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

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