nyoj----522 Interval （简单树状数组）

Interval

时间限制：2000 ms | 内存限制：65535 KB

难度：4

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

样例输出

上传者ACM_赵铭浩代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 #define maxn 200005   //整体平移100001个单位
 5 #define lowbit(x) ((x)&(-x))
 6 int aa[maxn+5];
 7 void ope(int x,int val)
 8 {
 9      x+=100001;
10     while(x<=maxn)
11     {
12         aa[x]+=val;
13         x+=lowbit(x);
14     }
15 }
16 long long getsum(int x)
17 {
18    long long ans=0;
19     while(x>0)
20     {
21         ans+=aa[x];
22         x-=lowbit(x);
23     }
24     return ans;
25 }
26 int main()
27 {
28     int test,nn,m,i,a,b;
29     scanf("%d",&test);
30     while(test--)
31     {
32       scanf("%d%d",&nn,&m);
33       memset(aa,0,sizeof(aa));
34       for(i=0;i<nn;i++)
35       {
36        scanf("%d%d",&a,&b);
37        ope(a,1);
38        ope(b+1,-1);
39       }
40       for(i=0;i<m;i++)
41       {
42        scanf("%d",&a);
43        a+=100001;
44        printf("%I64dn",getsum(a));
45       }
46     }
47   return 0;
48 }