acm 总是Time Limit Exceed！！！！！！！！！！！

Question

Description

The 36th ACM match was hosted in Chengdu in 2011 Nov 4. There is an interesting rule in ACM match that once you solve a problem, you'll get a balloon.

To simplify these names, we just name these teams S1, S2, S3 ... SN.

We want to know how many balloons they get between team Si and Sj (including Si and Sj). We may ask many times. It is too boring to count every time.

Since you are a programmer, can you write a program to solve it?

Input

First is an integer N (1 <= N <= 140) indicates the number of schools.

Followed with an integer M (1 <= M <= 50000) indicates the number of operations.

There are two kinds of operations.

AC X

     Means team SX solve a problem and will get a balloon.

COUNT X Y

     Means count balloons of team between SX and SY

Output

For each operation "COUNT X Y" output the number you count

Sample Input
5 4
AC 1
COUNT 1 3
AC 2
COUNT 1 3
Sample Output
1
2

我的程序：
#include
#include
#include
using namespace std;
int main()
{
int b,a;
scanf("%d%d",&b,&a);
int *c=new int[b];
for(int j=0;j {
c[j]=0;
}
while(a--)
{
string d;
int g,p,f;
cin>>d;
if(d=="AC")
{
scanf("%d",&g);
c[g-1]++;
}else if(d=="COUNT")
{
scanf("%d%d",&p,&f);
int sum=0;
for(int i=p-1;i<f;i++)
{
sum+=c[i];
}
printf("%d\n",sum);
}

}
return 0;
}

Answer 1

反正闲着没事，就去你们矿业大学的ACM注册了个账户，用java写了一个，你可以参考一下

import java.util.*;

public class Main {
    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);

        while (in.hasNext()) {
            int schools = in.nextInt();// 学校数
            int operations = in.nextInt();// 操作数
            int[] arr = new int[schools + 1];// 数组
            for (int i = 0; i < operations; i++) {
                String oper = in.next();
                if (oper.equals("AC")) {
                    int school = in.nextInt();
                    // 从当前学校往后的所有学校都加一
                    for (int j = school; j < schools + 1; j++) {
                        arr[j]++;
                    }
                } else {
                    int first = in.nextInt();
                    int second = in.nextInt();
                    System.out.println(arr[second] - arr[first - 1]);
                }
            }
        }
    }
}