Light oj 1030 二分查找

1088 - Points in Segments

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Given n points (1 dimensional) and q segments, you have to find the number of points that lie in each of the segments. A point p_i will lie in a segment A B if A ≤ p_i ≤ B.

For example if the points are 1, 4, 6, 8, 10. And the segment is 0 to 5. Then there are 2 points that lie in the segment.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 10⁵) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers denoting the points in ascending order. All the integers are distinct and each of them range in [0, 10⁸].

Each of the next q lines contains two integers A_k B_k (0 ≤ A_k ≤ B_k ≤ 10⁸) denoting a segment.

Output

For each case, print the case number in a single line. Then for each segment, print the number of points that lie in that segment.

Sample Input	Output for Sample Input
1 5 3 1 4 6 8 10 0 5 6 10 7 100000	Case 1: 2 3 2

Note

Dataset is huge, use faster I/O methods.

题目大意：数据量非常大，直接搞肯定就搞死了，明显二分查找，调用lower__bound和upper__bound就可以。

输出注意要用标准输入输出，否则超时gg.

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int maxn=100000+100;
int a[maxn];
int main()
{
    int kase=0;
    int T;
    int n,p;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&p);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
      printf("Case %d: ",++kase);
        while(p--)
        {
            int c,d;
            scanf("%d%d",&c,&d);
            int t1=upper_bound(a,a+n,d)-a;
            int t2=lower_bound(a,a+n,c)-a;
            printf("%d ",t1-t2);
        }
    }
    return 0;
}