SDUT oj 2610

 /*题目大意：输入一序列n个数字，然后输入m个询问，每个询问包含左边区间和右边区间，还有a和b，问你这个区间内有几个数大于等于a且小于等于b
 做法：树状数组，先求出这个区间内有几个数小于a，然后求这个区间内有几个数小于等于b，拿后者减去前者就是答案了*/
 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;
 const int maxn=;
 int c[maxn],ansl[maxn],ansr[maxn];
 struct pp
 {
     int num;
     int id;
 }p[maxn];
 struct p2p
 {
     int hi;
     int id;
     int l,r;
     int low;
 }p1[maxn];
 int cmp(pp a,pp b)
 {
         return a.num<b.num;
 }
 int cmp1(p2p a,p2p b)
 {
     return a.low<b.low;
 }
 int cmp2(p2p a,p2p b)
 {
     return a.hi<b.hi;
 }
 int n,m;
 int lowbit(int x)
 {
     return x&(-x);
 }
 void add(int x,int d)
 {
     while(x<=n)
     {
         c[x]+=d;
         x=x+lowbit(x);
     }
 }
 int sum(int x)
 {
     int ans=;
     while(x>)
     {
         ans+=c[x];
         x=x-lowbit(x);
     }
   return ans;
 }
 int main()
 {
     int i,j,k;
     int t;
     int cas=;
     scanf("%d",&t);
     while(t--)
     {
         memset(c,,sizeof(c));
         memset(ansl,,sizeof(ansl));
         scanf("%d%d",&n,&m);
         for(i=;i<=n;i++)
         {
             scanf("%d",&p[i].num);
             p[i].id=i;
         }
         for(i=;i<=m;i++)
         {
             scanf("%d%d%d%d",&p1[i].l,&p1[i].r,&p1[i].low,&p1[i].hi);
             p1[i].id=i;
         }
         sort(p+,p+n+,cmp);
         sort(p1+,p1+m+,cmp1);
         i=;
         j=;
        while(j<=m)
       {
           while(i<=n)
           {
             if(p[i].num>=p1[j].low)
             break;
             add(p[i].id,);
             i++;
           }
          while(j<=m)
          {
           if(i<=n&&p[i].num<p1[j].low)
           break;
           ansl[p1[j].id]=sum(p1[j].r)-sum(p1[j].l-);
           j++;
          }
       }
       i=;
       j=;
       memset(c,,sizeof(c));
       memset(ansr,,sizeof(ansr));
       sort(p1+,p1+m+,cmp2);
       while(j<=m)
       {
           while(i<=n)
           {
             if(p[i].num>p1[j].hi)
             break;
             add(p[i].id,);
             i++;
           }
          while(j<=m)
          {
           if(i<=n&&p[i].num<=p1[j].hi)
           break;
           ansr[p1[j].id]=sum(p1[j].r)-sum(p1[j].l-);
           j++;
          }
       }
       printf("Case #%d:\n",++cas);
       for(i=;i<=m;i++)
           if(ansr[i]-ansl[i]>)
           printf("%d\n",ansr[i]-ansl[i]);
           else printf("0\n");
     }
     return ;
 }