二分法应用--light oj1088 Points in Segments

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[666666];
int twofen(int *a,int k,int high)
{ if(k>a[high-1]) return high;
else{
    int low=0;high--;
    while(low<=high)
    {
        int mid=(low+high)/2;
        if(a[mid]<k)
        {
            low=mid+1;
        }
        else high=mid-1;
    }return low;}
}
int twofen2(int *a,int k,int high)
{ if(k>a[high-1]) return high;
else{
    int low=0;high--;
    while(low<=high)
    {
        int mid=(low+high)/2;
        if(a[mid]<=k)
        {
            low=mid+1;
        }
        else high=mid-1;
    }return low;}
}
int main()
{   int n,q,w,i,k,start,endd,flag=1;
    scanf("%d",&n);
    while(n--)
    {   ;int cal,low,high;
    printf("Case %d:\n",flag);flag++;
        scanf("%d%d",&q,&w);
        for(i=0;i<q;i++) scanf("%d",&a[i]);
        while(w--){
        scanf("%d%d",&start,&endd);
        low=twofen(a,start,q);//printf("%d\n",low);
        high=twofen2(a,endd,q);// printf("%d\n",high);
        printf("%d\n",high-low);
     }

    }

}

用二分法找出一个数组里，从A到B的范围一共多少个数，练习二分法，第二个调用函数包括相等的情况。