HDU－4190B - Distributing Ballot Boxes

Today, besides SWERC'11, another important event is taking place in Spain which rivals it in importance: General Elections. Every single resident of the country aged 18 or over is asked to vote in order to choose representatives for the Congress of Deputies and the Senate. You do not need to worry that all judges will suddenly run away from their supervising duties, as voting is not compulsory. 
 The administration has a number of ballot boxes, those used in past elections. Unfortunately, the person in charge of the distribution of boxes among cities was dismissed a few months ago due to nancial restraints. As a consequence, the assignment of boxes to cities and the lists of people that must vote in each of them is arguably not the best. Your task is to show how efficiently this task could have been done. 
 The only rule in the assignment of ballot boxes to cities is that every city must be assigned at least one box. Each person must vote in the box to which he/she has been previously assigned. Your goal is to obtain a distribution which minimizes the maximum number of people assigned to vote in one box. 
 In the first case of the sample input, two boxes go to the fi rst city and the rest to the second, and exactly 100,000 people are assigned to vote in each of the (huge!) boxes in the most efficient distribution. In the second case, 1,2,2 and 1 ballot boxes are assigned to the cities and 1,700 people from the third city will be called to vote in each of the two boxes of their village, making these boxes the most crowded of all in the optimal assignment.

Input

The fi rst line of each test case contains the integers N (1<=N<=500,000), the number of cities, and B(N<=B<=2,000,000), the number of ballot boxes. Each of the following N lines contains an integer a i,(1<=a i<=5,000,000), indicating the population of the i th city. 
 A single blank line will be included after each case. The last line of the input will contain -1 -1 and should not be processed.

Output

For each case, your program should output a single integer, the maximum number of people assigned to one box in the most efficient assignment.

题意：给你n个城市，m个箱子，现在求最小的投票箱容纳器

分析：如果n和m相等，那么容量最小肯定为人口数最多的那个，如果n<m，必须把所有箱子用完，那么箱子越多，平均值越小。
可以用二分枚举人口数量作为投票箱的容量，如果枚举后所需要的箱子数目小于m个，说明容量还可以再小，如果箱子数大于m个，说明最小容量就在当前容量到最大容量值之间

代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int s[5005000],n,m;
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==-1&&m==-1)
            break;
        int maxx=-1;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&s[i]);
            maxx=max(maxx,s[i]);    //找最大的人口数
        }
        int l=0,r=maxx,t;
        while(l<=r)                 //枚举
        {
            int mid=(l+r)/2;        //平均容量
            t=0;
            for(int i=0;i<n;i++)    //遍历一遍
            {
                if(mid>=s[i])       //如果该容量对该城市可行，+1
                    t+=1;
                else
                    t+=(s[i]+mid-1)/mid;  //该容量对该城市不够，那么+2
            }
            if(t>m)            //如果枚举出来箱子大于m个
                l=mid+1;
            else                //箱子数小于m个
                r=mid-1;
        }
        printf("%d\n",l);
    }
}