HDU4190Distributing Ballot Boxes

Distributing Ballot Boxes
Time Limit : 20000/10000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 2   Accepted Submission(s) : 2
Problem Description
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[sub]i[/sub],(1<=a[sub]i[/sub]<=5,000,000), indicating the population of the i[sup]th[/sup] 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.
 
Sample Input

2 7
200000
500000

4 6
120
2680
3400
200

-1 -1

 
Sample Output

100000
1700

 
Source
SWERC 2011

题意：将b个投票箱分放到n个城市，每个城市的人数不同，问箱子里最多可以装多少人的票

思路，，箱里至少有（总人数/总箱数）个票，最多为某最多人口城市的人数，然后用二分

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
int a[500005];
using namespace std;
int main()
{
    int n,b;
    int max,sum,num,evg;
    while(~scanf("%d%d",&n,&b))
    {

        if(n==-1&&b==-1)
        break;
        max=-1;
        num=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            num+=a[i];
            if(a[i]>max)
            max=a[i];
        }
        int left=num/b;
        int right =max;
        int mid;
        while(left<=right)
        {
            sum=0;
            mid=(left+right)/2;
            for(int i=0;i<n;i++)
            {
                if(a[i]<=mid)
                sum++;
                else
                sum=sum+(a[i]+mid-1)/mid;
                if(sum>b)
                break;
            }
            if(sum>b)    //if(sum>=b)结果会偏大，，因为要求的是最小满足条件的所以当刚好满足的时候应在较小的范围内找
            left=mid+1;
            else
            right=mid-1;
        }
        printf("%d
",left);
    }
    return 0;
}