Acdreamoj1108 (the kth number)

Question link: http://acdream.info/problem? PID = 1, 1108

Question: N number series. The number of occurrences of the number k in a certain interval in M queries. N, m <= 100000

Solution: because it is offline, it can be processed in a unified manner before being output. You can maintain a left and right pointers. Pre, pre [I] indicates the number of occurrences in the interval greater than or equal to I. To reduce the complexity, the key is the moving method of left and right, that is, how to sort the query interval. If the range is sorted by the Left endpoint, the maximum return of the right endpoint is N at a time, if you sort by the right endpoint, the left endpoint must be n at most each time. There is a good solution here, that is, fuzzy sorting. The left endpoint is not strictly ordered, that is, divided by SQRT (N) and then sorted. In this way, the maximum complexity is N * SQRT (n)

Code:

/******************************************************* @author:xiefubao*******************************************************/#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <vector>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <string.h>//freopen ("in.txt" , "r" , stdin);using namespace std;#define eps 1e-8#define zero(_) (abs(_)<=eps)const double pi=acos(-1.0);typedef long long LL;const int Max=100010;const int INF=1e9+7;int num[Max];int help[Max];int r[Max];int l[Max];int k[Max];int pre[Max];int cnt[Max];int tool;bool cmp(int i,int j){    if(l[i]/tool==l[j]/tool&&r[i]!=r[j])        return r[i]<r[j];    return l[i]<l[j];}int ans[Max];int n,m;int findans(int t){    int l=1,r=n;    while(l<=r)    {        int middle=(l+r)/2;        if(pre[middle]>=t)        l=middle+1;        else            r=middle-1;    }    return l-1;}int main(){    int t;    cin>>t;    while(t--)    {        scanf("%d%d",&n,&m);        tool=sqrt(n);        for(int i=0; i<n; i++)            scanf("%d",num+i),help[i]=i;        for(int i=0; i<m; i++)            scanf("%d%d%d",l+i,r+i,k+i),l[i]--,r[i]--;        sort(help,help+m,cmp);        memset(cnt,0,sizeof cnt);        memset(pre,0,sizeof pre);        int left=0,right=-1;        for(int i=0;i<m;i++)        {           int L=l[help[i]],R=r[help[i]];           while(left<L){ pre[cnt[num[left++]]--]--;}           while(L<left){ pre[++cnt[num[--left]]]++;}           while(right<R){ pre[++cnt[num[++right]]]++;}           while(R<right){ pre[cnt[num[right--]]--]--;}           ans[help[i]]=findans(k[help[i]]);        }        for(int i=0;i<m;i++)            printf("%d\n",ans[i]);    }    return 0;}