poj 3258 River Hopscotch

#include <iostream>            //二分法，参照poj 3273
#include <algorithm>
using namespace std;
int main()
{
    int list[50010],L,N,M,i;
    scanf("%d%d%d",&L,&N,&M);
    list[0]=0;list[N+1]=L;
    for(i=1;i<=N;++i)
    {
        scanf("%d",&list[i]);
    }
    sort(list,list+N+2);
    int l=0,r=L,mid,res;
    while(l<r)
    {
        int count=0,s=0;
        mid=l+(r-l+1)/2;        //不是mid=(l+r)/2;也不是mid=l+(r-l)/2;
//这一步至关重要,如l=3,r=4,mid=3,会一直执行l=mid,无限循环,而mid=l+(r-l+1)/2;保证了每次待查找序列的规模都会变小
        for(i=1;i<=N+1;++i)
        {
            if(s+list[i]-list[i-1]>=mid)
                s=0;
            else
            {
                count++;
                s+=list[i]-list[i-1];
            }    
            if(count>M)
                break;
        }
        if(count>M)
            r=mid-1;
        else 
            l=mid;
    }
    printf("%d\n",l);
    return 0;
}

//对应于 1100序列中最右侧的1的查找（最大值）:
//http://duanple.blog.163.com/blog/static/709717672009049528185/