二分练习

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

输入

输出

示例输入

8 4
1 2 3 4 5 6 8 11
4
9
2
7

示例输出

4
8
2
6 8

#include <stdio.h>
int a[100000010];
void qsort(int *a,int star ,int fin)//快排函数
{
    int key=a[star],i=star,j=fin;
    if(star>=fin)
        return;
    while(i<j)
    {
        while(i<j&&a[j]>=key) j--; a[i]=a[j];
        while(i<j&&a[i]<=key) i++;
        a[j]=a[i];
    }
    a[i]=key;
    qsort(a,star,i-1);
    qsort(a,i+1,fin);
}
int down(int *a,int low,int high,int key)//寻找key的前一个数
{
    int mid;
    int r=-1;
    while(low<=high)
    {
        mid=(low+high)/2;
        if(a[mid]<=key)
        {
            low=mid+1;
            r=mid;
        }
        else high=mid-1;
    }
    return r;
}
int up(int *a,int low,int high,int key )//寻找key的后一个数
{
    int mid;
    int r=-1;
    while(low<=high)
    {
        mid=(low+high)/2;
        if(a[mid]>=key)
        {
            r=mid;
            high=mid-1;
        }
        else
            low=mid+1;
    }
    return r;
}
int main()
{
    int n,m,i,k;
    while(~scanf("%d %d",&n,&m))
    {
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        qsort(a,0,n-1);
        for(i=0; i<m; i++)
        {
            scanf("%d",&k);
            int kmax=up(a,0,n-1,k);
            int kmin=down(a,0,n-1,k);
            if(kmax==-1)//若不存在比key大的，输出比key小的
                printf("%d\n",a[kmin]);
            else if(kmin==-1)//若不存在比key小的，输出比key大的
                printf("%d\n",a[kmax]);
            else if(a[kmax]==a[kmin])
                printf("%d\n",a[kmin]);
            else
            {
                if((a[kmax]-k)==(k-a[kmin]))
                    printf("%d %d\n",a[kmin],a[kmax]);
                else if((a[kmax]-k)>(k-a[kmin]))
                    printf("%d\n",a[kmin]);
                else
                    printf("%d\n",a[kmax]);
            }
        }
        printf("\n");
    }
    return 0;
}