SRM 586 DIV2 500

Problem Statement
	F is a function that is defined on all real numbers from the closed interval [1,N]. You are given a vector <int>Y with N elements. For each i (1 <= i <= N) we have F(i) =Y[i-1]. Additionally, you know that F is piecewise linear: for each i, on the interval [i,i+1] F is a linear function. The function F is uniquely determined by this information. For example, if F(4)=1 and F(5)=6 then we must have F(4.7)=4.5. As another example, this is the plot of the function F for Y = {1, 4, -1, 2}. You are also given a vector <int> query. For each i, compute the number of solutions to the equation F(x) =query[i]. Note that sometimes this number of solutions can be infinite. Return a vector <int> of the same length as query. For each i, element i of the return value should be -1 if the equation F(x) =query[i] has an infinite number of solutions. Otherwise, element i of the return value should be the actual number of solutions this equation has.
Definition
	Class: PiecewiseLinearFunctionDiv2 Method: countSolutions Parameters: vector <int>, vector <int> Returns: vector <int> Method signature: vector <int> countSolutions(vector <int> Y, vector <int> query) (be sure your method is public)
Constraints
-	Y will contain between 2 and 50 elements, inclusive.
-	Each element of Y will be between -1,000,000,000 and 1,000,000,000, inclusive.
-	query will contain between 1 and 50 elements, inclusive.
-	Each element of query will be between -1,000,000,000 and 1,000,000,000, inclusive.
Examples
0)
	{1, 4, -1, 2} {-2, -1, 0, 1} Returns: {0, 1, 2, 3 } This is the example from the problem statement. The detailed information about the queries is: There is no such x that F(x) = -2 is satisfied. F(x) = -1 is only true for x = 3. F(x) = 0 has two roots: 2.8 and 10/3. F(x) = 1 has three roots: 1, 2.6 and 11/3.
1)
	{0, 0} {-1, 0, 1} Returns: {0, -1, 0 } This function's plot is a horizontal segment between points (1, 0) and (2, 0). F(x) = 0 is satisfied for any x between 1 and 2 and thus the number of solutions is infinite. For any other value on the right-hand side, it has no solutions.
2)
	{2, 4, 8, 0, 3, -6, 10} {0, 1, 2, 3, 4, 0, 65536} Returns: {3, 4, 5, 4, 3, 3, 0 }
3)
	{-178080289, -771314989, -237251715, -949949900, -437883156, -835236871, -316363230, -929746634, -671700962} {-673197622, -437883156, -251072978, 221380900, -771314989, -949949900, -910604034, -671700962, -929746634, -316363230} Returns: {8, 6, 3, 0, 7, 1, 4, 8, 3, 4 }

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题目意思:

两个vector,  Y 和 query

Y代表的是 纵坐标,(横坐标为 i+1); (如上图所示)

query代表查询;  询问针对每个query[i] 能在那些线性区间中找到几个点.满足 纵坐标为query[i],即点(x,query[i])

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class PiecewiseLinearFunctionDiv2 
{
public:
	vector <int> countSolutions(vector <int>, vector <int>);
};

vector <int> PiecewiseLinearFunctionDiv2::countSolutions(vector <int> Y, vector <int> query) 
{
    vector <int> ans;
    int num;
	for(int i = 0; i < query.size(); ++i)
	{
	    num = 0;
	    for(int j = 0; j < Y.size()-1; ++j)
	    {
	        if(Y[j]==Y[j+1] && query[i]==Y[j])
            {
                num = -1;
                break;
            }
	        if(min(Y[j],Y[j+1]) < query[i] && query[i] < max(Y[j+1],Y[j]))
                num++;
            if(query[i] == Y[j])
                num++;
	    }
	    if(num!=-1 && query[i]==Y[Y.size()-1])
            num++;
	    ans.push_back(num);
	}
	return ans;
}

<%:testing-code%>
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