zoj 2132 the most frequent number

the  most frequent number

CE: Gcc注释 用/**/

MLE: 快排 取中位数

/*zoj 2132 the most frequent number
 quick sort */
# include <stdio.h>
# include <stdlib.h>
# define MAXN 250005

int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}
int main()
{
    int a[MAXN];
    int L, i;
    while (scanf("%d", &L)==1)
    {
        for (i=0; i<L; ++i)
            scanf("%d", &a[i]);
        qsort(a, L, sizeof(int), compare);
        printf("%d\n",a[L/2]);
    }
    return 0;
}

看了别人的代码，

/*
from 百度空间 kobeddshow
http://hi.baidu.com/kobeddshow/blog/item/26bc91ad166ba09246106439.html
*/
# include<stdio.h>
int main()
{
    int n,a,i,ans,k;
    while(scanf("%d",&n) != EOF )
    {
        k = 0;
        for(i=0; i<n; i++)
        {
            scanf("%d",&a);
            if (k == 0) ans=a, k=1;
            else if(ans == a) ++k;
            else --k;
        }
        printf("%d\n",ans);
    }
return 0;  /*没有这一行，提示程序非零退出 */
}

这个方法充分利用了要求的数比总数一半还多这个条件：只有比一半多，才能保证保留要求的ans，比一半少一个都不行。
看书算法技巧方面的不够多，做题也是刚开始，要坚持不懈！

TLE: 自定义快排 取中位数

View Code

/*zoj 2132 the most frequent number
 quick sort */
# include <stdio.h>
# include <stdlib.h>
# define MAXN 250005
void quick_sort(int *a, int left, int right);
int main()
{
    int a[MAXN];
    int L, i;
    while (scanf("%d", &L)==1)
    {
        for (i=1; i<=L; ++i)
            scanf("%d", &a[i]);
        quick_sort(a, 1, L);
        printf("%d\n",a[L/2+1]);
    }
    return 0;
}
void quick_sort(int *a, int left, int right)
{
    int tmp, i, last;
    if (left >= right) return;
    tmp = a[left];
    a[left] = a[(left+right)>>1];
    a[(left+right)>>1] = tmp;
    last = left;
    for (i = left+1; i <= right; ++i)
        if (a[i]<a[left] && (++last)!=i)        /*短路*/
        {
            tmp = a[last];
            a[last] = a[i];
            a[i] = tmp;
        }
    tmp = a[last];
    a[last] = a[left];
    a[left] = tmp;
    quick_sort(a, left, last-1);
    quick_sort(a,last+1, right);
}