Lightoj 1088

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

题意： 有一维的n个点和q条线段。询问每条线段上的点有多少个。

思路：寻找这些点中对于每条线段的上下界就可以。

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

int n, q;
int a[100010];
int x, y;
int ok;

int query(int t,int &ok)
{
    ok = 0;
    int left = 1;
    int right = n;
    int mid;
    while (left <= right)
    {
        int mid = (left + right) / 2;
        if (a[mid] >= t)
        {
            if (a[mid] == t)
                ok = 1;
            right = mid - 1;
        }
        else
            left = mid + 1;
    }
    return left;
}

int main()
{
    int t;
    scanf("%d",&t);
    int cases = 1;
    while (t--)
    {
        scanf("%d%d",&n,&q);
        for (int i = 1; i <= n; i++)
            scanf("%d",&a[i]);
        printf("Case %d:
", cases++);
        while (q--)
        {
            scanf("%d%d", &x, &y);
            int ok1 = 0, ok2 = 0, tmp = 0;
            int t1 = query(x,ok1);
            int t2 = query(y,ok2);
            if (ok2) tmp = 1;
            //printf("      %d %d
",t1,t2);
            printf("%d
", t2 - t1 + tmp);
        }
    }
    return 0; 
}