A, subject to the effect
subject portal
finds the closest element (divide and conquer / binary search): In a non-reduced sequence, and find the closest value given element.
[Enter]
The first row contains an integer n, a non-drop sequence length.
The second line contains n integers, the non-descending sequence of elements.
The third row contains an integer m, to ask for the number of a given value. 1 <= m <= 10000.
Next m lines, each an integer, the closest element to ask for the given value. All values are given size between 0-1,000,000,000
[Output]
m lines of an integer value closest to the corresponding element of a given value, holding the input order. If the minimum values of a plurality of conditions are satisfied, the output
input：
3
2 5 8
2
10
5
output：
8
5
Ideas are as follows:
The problem is to use the binary method is closest to [] array value m at a given, because it is the closest that we need to conduct a more detailed discussion, here I will have a solution to this problem in two ways:
the first way is to write a recursive half ( 但是注意这种方法在这一题是行不通的因为数据太大，递归层数太多，造成栈溢出,但是分类讨论的思想还是值的我们学习的); another method is to write a half while loop
Recursive solution to a problem
/*二分递归找最近的值的步骤：
主要思想是：在不断细分区间的时候，通过不同的判断条件，去不断的选择符合题意的区间，当选择了某一个区间后，并对该区间进行分类讨论：讨论1.看min与边界的差值是否为1，如果是1，那么我们想要的结果就是在ar[mid]与ar[边界值]之间选择（通过看值两个数与所要找的数的差值，进行比较，还要注意两者均是可以选择的）
1.讨论要查找的值是否在所求区间的外边（那么最接近要查找的值就是边界值，所以接下来要对边界值进行讨论）
2.讨论 要查找的值位于做边界值与ar[mid] 之间
3.讨论 要查找的值位于ar[mid]与右边界值之间
4.讨论 要查找的值等与ar[mid]的情况
*/
#include<stdio.h>
#include<stdlib.h>
int a[10005];//存放升序排列的一列数字
int b[10005];//存放要找的数字
int m;//输入要查找的数的个数
long long finder(int mx, int mn, int mid, int j)//
{//第一步区分查找的这个数是否在开区间内
if (b[j] <= a[mn] || b[j] >= a[mx])//不在
{
//对内部进行细分，到底是比最小的还小，还是比最大的还大
if (b[j] <= a[mn]) //比最小的还小
printf("%d\n", a[mn]);
else//比最大的还大
printf("%d\n", a[mx]);
}
else//位于最小的数字和最大的数字之间
{
//根据mid来分，比mid大还是比mid小，还是和mid相等
if (a[mn]<b[j] && b[j]<a[mid])//比mid小
{
if (mid - mn != 1)
{
mx = mid - 1;//将mx替换为mid
mid = (mn + mx) / 2;//重新计算mid下标
finder(mx, mn, mid, j);
}
else
{
if (a[mid] - b[j]>b[j] - a[mn])//比较距离
printf("%d\n", a[mn]);
else if (a[mid] - b[j]<b[j] - a[mn])
printf("%d\n", a[mid]);
else
printf("%d %d\n", a[mn], a[mid]);
}
}
else if (a[mid]<b[j] && b[j]<a[mx])//比mid大
{
if (mx - mid != 1)
{
mn = mid;//重新计算mn
mid = (mn + mx) / 2;//重新计算mid下标
finder(mx, mn, mid, j);
}
else
{
if (b[j] - a[mid]>a[mx] - b[j])
printf("%d\n", a[mx]);
else if (b[j] - a[mid]<a[mx] - b[j])
printf("%d\n", a[mid]);
else
printf("%d %d\n", a[mid], a[mx]);
}
}
else//和mid相等
printf("%d\n", b[j]);
}
return 0;
}
int main()
{
int j = 0;//用作函数循环的
int n;//输入非降序序列的长度
scanf("%d", &n);
for (int i = 0; i<n; i++) //循环输入n个数字
scanf("%d", &a[i]);
scanf("%d", &m); //输入要查询的数字个数
for (int i = 0; i<m; i++) //循环输入要查找的数，存在b[i]中
scanf("%d", &b[i]);
int mx = n - 1, mn = 0;
int mid = (mx + mn) / 2;
for (j = 0; j<m; j++)//循环调用函数，把b[j]中存储的所有要找的数字找完
finder(mx, mn, mid, j);
return 0;
}
whille circulation problem solution
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
long long a[100001];
int n;
long long Binary_search(long long int b)
{
//首先对是否在该区间以外进行讨论
int l = 0;
int r = n - 1;
if(b <= a[l])
return a[l];
if(b >= a[r])
return a[r];
//在所求区间内进行二分讨论
while(r - l > 1) //while循环里面的r - l > 1 这个限定条件很重要，当r yu l 的差值为1 的时候就应该停止循环了，在分别ar[l] 与 ar[r] 两个值讨论看那个值更合适
{
int mid = (l + r) / 2;
if(b >= a[mid])
l = mid; //在这里 l 直接等于 mid ，否则会出错
else
r = mid; // 同理
}
long long int x = abs(a[l] - b);
long long int y = abs(a[r] - b);
return x <= y ? a[l] : a[r];
}
int main()
{
// freopen("T.txt","r",stdin);
int i,m;
long long b;
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%lld",&a[i]);
}
scanf("%d",&m);
while(m){
scanf("%lld",&b);
printf("%lld\n",Binary_search(b));
m--;
}
return 0;
}
