#include <iostream>#include <stdio.h>#include <vector>using namespace std;/*問題:Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.You may assume no duplicates in the array.Here are few examples.[1,3,5,6], 5 → 2[1,3,5,6], 2 → 1[1,3,5,6], 7 → 4[1,3,5,6], 0 → 0分析:這是二分查找的lowwer_bound的問題。輸入:4 51 3 5 64 21 3 5 64 71 3 5 64 01 3 5 6輸出2140關鍵:1 lowwer_bound:		//low == high時，如果找到，就返回		if(nums.at(low) >= target)		{			return low;		}		//說明是數組最后一個元素，返回low+1		else		{			return low + 1;		}*/class Solution {public:	int lower_bound(vector<int>& nums , int target)	{		if(nums.empty())		{			return -1;		}		int low = 0;		int high = nums.size() - 1;		int mid;		while(low < high)		{			mid  = low + (high - low) / 2;			//中間大于目標值，目標值，mid可能是結果,繼續在左半部分尋找			if(nums.at(mid) >= target)			{				high = mid;			}			//中間值 < 目標值，mid不可能是結果，在右半部分尋找			else			{				low = mid + 1;			}		}		//low == high時，如果找到，就返回		if(nums.at(low) >= target)		{			return low;		}		//說明是數組最后一個元素，返回low+1		else		{			return low + 1;		}	}    int searchInsert(vector<int>& nums, int target) {		int high = lower_bound(nums , target);		return high;    }};void PRocess(){	int num ;	int value;	vector<int> nums;	int target;	Solution solution;	vector<int> results;	while(cin >> num >> target)	{		nums.clear();		for(int i  = 0 ; i < num ; i++)		{			cin >> value;			nums.push_back(value);		}		int result = solution.searchInsert(nums , target);		cout << result << endl;	}}int main(int argc , char* argv[]){	process();	getchar();	return 0;}