NYOJ 522 裸的树状数组

       又是一道水题，，话说这次月赛水题真的很多很多，貌似比赛时写出来的题都是水题。。。看来，水平也就能水一下题而已。。。。不过这道题比赛时还是坑了不少人，很多人在处理边界问题0的时候没有注意，都TLE了，，当时我也TLE了一次，后来仔细想了想，改过后就ac了。相比那些一直TLE到最后的孩纸来说，我算是幸运了。不过，这道题难度有点高了，除了边界问题外，就是道裸的树状数组，没什么难度的。题目：

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

23 41 31 22 301231 30 0-101

样例输出

0232010

ac代码：

 #include <iostream>#include <cstdio>#include <string.h>#include <cmath>const int N=200002;const int M=100001;int num[N],ans[N];int lowbit(int x){return x&(-x);}void update(int m,int value){while(m>0){  num[m]+=value;  m-=lowbit(m);}}int sum(int x){int s=0;while(x<=N){  s+=num[x];  x+=lowbit(x);}return s;}int main(){//freopen("1.txt","r",stdin);int numcase;scanf("%d",&numcase);while(numcase--){  memset(num,0,sizeof(num)); // memset(ans,0,sizeof(ans));  int n,m;  scanf("%d%d",&n,&m);  int x,y;  while(n--){    scanf("%d%d",&x,&y);x=x+M;y=y+M;update(x-1,-1);update(y,1);  }  while(m--){    scanf("%d",&x);x=x+M;//if(!ans[x])  printf("%d\n",sum(x));/*elseprintf("%d\n",ans[x]);*/  }}return 0;}