//HDOJ 4417 Super Mario 线段树 转化为单点更新 成段查询/*题意：查询区间内小于等于某数的数的总和思路：离线查询，先记录所有的询问并排序，然后一边删一边查，查之前把不满足的删掉 就转化为简单的单点更新，成段求和。*/#include<stdio.h>#include<string.h>#include<stdlib.h>#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define N 100005int T,n,m,cnt,ca = 1;int sum[N<<2],ans[N];struct Node{int x;int id;}s[N];struct node{int from;int to;int k;int id;}q[N];int cmp(const void *a,const void *b){return (*(Node *)b).x - (*(Node *)a).x;}int cmp1(const void *a,const void *b){return (*(node *)b).k - (*(node *)a).k;}void Pushup(int rt){sum[rt] = sum[rt<<1]+sum[rt<<1|1];}void Debug(int n){int i;for(i = 1; i <= n; ++i)printf("sum[%d] : %d\n",i,sum[i]);}void Build(int rt,int l,int r){if(l == r){scanf("%d",&s[cnt].x);s[cnt].id = cnt++;sum[rt] = 1;return ;}int mid = (l + r) >> 1;Build(lson);Build(rson);Pushup(rt);}void Update(int rt,int l,int r,int x,int c){if(l == r){sum[rt] += c;return ;}int mid = (r + l) >> 1;if(x <= mid)Update(lson,x,c);elseUpdate(rson,x,c);Pushup(rt);}int Query(int rt,int l,int r,int L,int R){if(L <= l && R >= r){return sum[rt];}int res = 0;int mid = (l + r) >> 1;if(L <= mid) res += Query(lson,L,R);if(R > mid ) res += Query(rson,L,R);return res;}void Init(){int i;cnt = 1;scanf("%d %d",&n,&m);Build(1,1,n);qsort(s+1,n,sizeof(s[0]),cmp);for(i = 1; i <= m; ++i){scanf("%d %d %d",&q[i].from,&q[i].to,&q[i].k);q[i].id = i;}qsort(q+1,m,sizeof(q[0]),cmp1);}void Solve(){int i,j = 1;memset(ans,0,sizeof(ans));for(i = 1; i <= n; ++i){while(s[i].x <= q[j].k){ans[q[j].id] = Query(1,1,n,q[j].from+1,q[j].to+1);++j;if(j > m)break;}if(s[i].x > q[j].k){Update(1,1,n,s[i].id,-1);}}}void Print(){int i;printf("Case %d:\n",ca++);for(i = 1; i <= m; ++i)printf("%d\n",ans[i]);}int main(){scanf("%d",&T);while(T--){Init();Solve();Print();}return 0;}
