nyist 522 Interval

描述

There are n(1 <= n <= 100000) intervals [ai, bi] and m(1 <= m <= 100000) queries, -100000 <= ai <= bi <= 100000 are integers.

Each query contains an integer xi(-100000 <= x <= 100000). For each query, you should answer how many intervals convers xi.

输入

The first line of input is the number of test case.
For each test case,
two integers n m on the first line,
then n lines, each line contains two integers ai, bi;
then m lines, each line contains an integer xi.

输出

m lines, each line an integer, the number of intervals that covers xi.

样例输入

2
3 4
1 3
1 2
2 3
0
1
2
3
1 3
0 0
-1
0
1

样例输出

0
2
3
2
0
1
0

这个题是一个树桩数组题 但是和常规不同的就是这里边有负数
把负数转换为正数来处理，

#include <stdio.h>
#include <string.h>

#define N 100000

int c[N*4];

int bity(int x)
{
    return x&(-x);
}

void add(int k,int num)
{

    while(k>0)
    {
        c[k] += num;
        k -= bity(k);
    }

}

int get_sum(int x,int n)
{
    int sum = 0;
    /*if(x ==0)
     return sum;*/
    // else
     {
    while(x<=n)
    {
       sum += c[x];
       x += bity(x);
    }
    return sum;
     }
}

int main()
{
    //printf("%d\n",bity(-1));
    int t;
    scanf("%d",&t);
    while(t--)
    {
           int n,m;
           scanf("%d%d",&n,&m);
           memset(c,0,sizeof(c));
           int i = 0;
           int a = 0,b = 0;

           for(i = 0; i < n; i++)
                 {
                      scanf("%d%d",&a,&b);
                      add(b+N+10,1);
                      add(a+N+10-1,-1);
                 }
           while(m--)
           {
              int x;
              scanf("%d",&x);
              printf("%d\n",get_sum(x+N+10,n+N*2+10));
           }

    }
    return 0;
}