LightOJ 1088 - Points in Segments （二分）

题目链接：http://lightoj.com/volume_showproblem.php?problem=1088

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.

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PROBLEM SETTER: JANE ALAM JAN

题目大意：n个数，然后m个区间，求各个区间在n个数中包含个数

题目解析：直接二分搜索即可

代码如下：

#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<vector>#include<set>#include<string>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#define N 100009using namespace std;const int inf = 1e9;const int mod = 1<<30;const double eps = 1e-8;const double pi = acos(-1.0);typedef long long LL;int a[N];int main(){    int n, i, k, t;    int s, e, cnt = 0;    cin >> t;    while(t--)    {        printf("Case %d:\n", ++cnt);        scanf("%d%d", &n, &k);        for(i = 0; i < n; i++) scanf("%d", &a[i]);        a[n] = inf;        for(i = 1; i <= k; i++)        {            scanf("%d%d", &s, &e);            int f = 0, ans = lower_bound(a, a + n + 1, e) - lower_bound(a, a + n + 1, s);            f = e == *lower_bound(a, a + n + 1, e) ? 1 : f;            if(f) ans++;  // 或者 ans = upper_bound(a, a + n, e) - lower_bound(a, a + n, s);            printf("%d\n", ans);        }    }    return 0;}