简单题练习——数字在排序数组中出现的次数

题目描述：

统计一个数字在排序数组中出现的次数。

输入：

每个测试案例包括两行：

第一行有1个整数n，表示数组的大小。1<=n <= 10^6。

第二行有n个整数，表示数组元素，每个元素均为int。

第三行有1个整数m，表示接下来有m次查询。1<=m<=10^3。

下面有m行，每行有一个整数k，表示要查询的数。

输出：

对应每个测试案例，有m行输出，每行1整数，表示数组中该数字出现的次数。

样例输入：

81 2 3 3 3 3 4 513

样例输出：

4

总结：卡了好久，总是超时，最后把最多1k的数值的输入输出都改成scanf printf才勉强通过，cin 效率差这么多吗（超过200ms）按理说相对1M的输入这里1K基本可以忽略呀，

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
int num[1000000];
 
int biSearchFirst(int n, int val)//第一个
{
    int low=0; 
    int high=n-1;
    while (low<high)
    {
        int mid=(low+high)/2;
        if(num[mid]<val)
            low = mid+1;
        else
            high = mid;
    }
    return low;
}
 
int biSearchLast(int n, int val)//第一个大于的
{
    int low=0; 
    int high=n-1;
    while (low<high)
    {
        int mid=(low+high)/2;
        if(num[mid]>val)
            high = mid;
        else
            low=mid+1;
        /*int mid=(low+high)/2;//弄了个死循环
        if(num[mid]>val)
            high = mid-1;
        else 
            low=mid;*/
    }
    if(high==n-1 && num[high]==val)
        high++;
     
    return high;
}
 
int OccursNum(int n, int val)
{
    int start = biSearchFirst(n,val);
    int end = biSearchLast(n,val);
    if(start!=-1 && end!=-1)
        return end-start/*+1*/;
    else
        return 0;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while (cin>>n)
    {
        for(int i=0; i<n; i++)
            scanf("%d",num+i);
        int m;
        scanf("%d",&m);
        //cin>>m;
        for(int i=0; i<m; i++)
        {
            int temp;
            scanf("%d",&temp);
            //cin>>temp;
            printf("%d\n",OccursNum(n,temp));
            //cout<<OccursNum(n,temp)<<endl;
        }
    }
    return 0;
}
/**************************************************************
    Problem: 1349
    User: xjbscut
    Language: C++
    Result: Accepted
    Time:1000 ms
    Memory:5416 kb
****************************************************************/

后面把k次查询排序组织按需查询缩小后面的查询范围效率也没有太大优化（一组简单测试样例没通过，有点小bug）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <vector>
using namespace std;
int num[1000000];
struct ansInfo
{
    int id;
    int queryNum;
    int val;
};
ansInfo Ans[1000];
int biSearchFirst(int start, int end, int val)
{
    int low=start; 
    int high=end;
    while (low<=high)
    {
        int mid=(low+high)/2;
        if(num[mid]>val)
            high = mid-1;
        else if(num[mid]<val)
            low = mid+1;
        else
        {
            if(mid>0 && num[mid-1]==val)
                high=mid-1;
            else
                return mid;
        }
    }
}
 
int biSearchLast(int start, int end, int val)
{
    int low=start; 
    int high=end;
    while (low<=high)
    {
        int mid=(low+high)/2;
        if(num[mid]>val)
            high = mid-1;
        else if(num[mid]<val)
            low = mid+1;
        else
        {
            if(mid<end && num[mid+1]==val)
                low=mid+1;
            else
                return mid;
        }
    }
    return -1;
}
 
//int OccursNum(int n, int val)
//{
//  int start = biSearchFirst(n,val);
//  int end = biSearchLast(n,val);
//  if(start!=-1 && end!=-1)
//      return end-start+1;
//  else
//      return 0;
//}
bool cmp1(const ansInfo &a, const ansInfo &b)
{
    return a.queryNum<b.queryNum;
}
bool cmp2(const ansInfo &a, const ansInfo &b)
{
    return a.id<b.id;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while (cin>>n)
    {
        for(int i=0; i<n; i++)
            scanf("%d",num+i);
        int m;
        scanf("%d",&m);
        //cin>>m;
        ansInfo temp;
        for(int i=0; i<m; i++)
        {
            temp.id=i;
            scanf("%d",&temp.queryNum);
            //cin>>temp.queryNum; 
            Ans[i]=temp;
        }
        sort(Ans,Ans+m,cmp1);
        int first,last;
        first=biSearchFirst(0,n-1,Ans[0].queryNum);
        last=biSearchLast(0,n-1,Ans[0].queryNum);
        Ans[0].val=last-first+1;
        for(int i=1; i<m; i++)
        {
            if(Ans[i].queryNum == Ans[i-1].queryNum)
                Ans[i].val=Ans[i-1].val;
            else
            {
                first=biSearchFirst(last+1,n-1,Ans[i].queryNum);
                last=biSearchLast(first,n-1,Ans[i].queryNum);
                Ans[i].val=last-first+1;
            }
        }
        sort(Ans,Ans+m,cmp2);
        for(int i=0; i<m; i++)
            printf("%d\n",Ans[i].val);
            //cout<<Ans[i].val<<endl;
    }
    return 0;
}