[poj3368]Frequent values

  题目传送门:http://poj.org/problem?id=3368

  题解传送门:http://www.cnblogs.com/frog112111/p/3306280.html

  这道题我又看了题解(捂脸),题解大家看链接的吧。还有就是我的代码较拙,跑了1s多。。。。。。

  

#include<cstdio>
#include<cstring>
#include<map>
#include<iostream>
using namespace std;
const int maxn=100010;
map<int,int> c;
int n,q,f[maxn][20],num[maxn];

void init()
{
     int i,j;
     for (j=1;(1<<j)<=n;j++)
         for (i=1;i+(1<<j)-1<=n;i++)
             f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);
}

int query(int l,int r)
{
    int k=0;
    if (l>r)return 0;
    while ((1<<(k+1))<=(r-l+1))k++;
    return max(f[l][k],f[r-(1<<k)+1][k]);
}
int main()
{
    while (scanf("%d",&n) && n)
    {
          int i,j,l,k,ans;
          scanf("%d",&q);
          c.clear();
          memset(f,0,sizeof(f));
          memset(num,0,sizeof(num));
          for (i=1;i<=n;i++)
          {
              scanf("%d",&num[i]);
              c[num[i]]++;  
              f[i][0]=c[num[i]];      
          }
          init();
          for (i=1;i<=q;i++)
          {
              scanf("%d%d",&j,&l);
              k=j;
              while (num[k]==num[k-1] && k<=l)k++;
              ans=query(k,l);
              if (k!=j)ans=max(k-j,ans);
              printf("%d\n",ans);
          }
    }
    return 0;
}

 

  

posted @ 2016-02-23 23:06  Sun_Sea  阅读(130)  评论(0编辑  收藏  举报