Day 39 A. Odd Selection

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该程序解决了一个数学问题，即在给定的一组整数中，判断是否能选取指定数量的元素，使得这些元素的和为奇数。通过计算数组中奇数和偶数的数量，利用奇偶性进行判断。当数量相等且等于目标取数时，如果奇数个数为偶数则不可能；其他情况下，只要存在奇数或者在数量大于目标时偶数不全选，都可以找到满足条件的子集。

Problem
Shubham has an array a of size n, and wants to select exactly x elements from it, such that their sum is odd. These elements do not have to be consecutive. The elements of the array are not guaranteed to be distinct.

Tell him whether he can do so.

Input
The first line of the input contains a single integer t (1≤t≤100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers n and x (1≤x≤n≤1000) — the length of the array and the number of elements you need to choose.

The next line of each test case contains n integers a1,a2,…,an (1≤ai≤1000) — elements of the array.

Output
For each test case, print “Yes” or “No” depending on whether it is possible to choose x elements such that their sum is odd.

You may print every letter in any case you want.

Example
input
5
1 1
999
1 1
1000
2 1
51 50
2 2
51 50
3 3
101 102 103

output
Yes
No
Yes
Yes
No

Note
For 1st case: We must select element 999, and the sum is odd.

For 2nd case: We must select element 1000, so overall sum is not odd.

For 3rd case: We can select element 51.

For 4th case: We must select both elements 50 and 51 — so overall sum is odd.

For 5th case: We must select all elements — but overall sum is not odd.

题目意思为给你n个数问在这n个数中取x个数是否能让这x个数总和加起来为奇数
//这里用到了奇偶性当n=x时如果有偶数个奇数那么是不成立的其他皆成立当n>x时如果取得数为偶数个并且没有偶数那么也是不成立

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stdio.h>
#include<limits.h>
#include<cmath>
#include<set>
#include<map>
#define ll long long
using namespace std;
int a[1005];
int main()
{
    int t;
    int n, x;
    int sum1, sum2;
    cin >> t;
    while (t--)
    {
        sum1 = 0;
        sum2 = 0;
        cin >> n;
        cin >> x;
        for (int i = 0; i < n; i++) 
        {
            cin >> a[i];
            if (a[i] % 2 == 1) 
            {
                sum1++;
            }
            else
            {
                sum2++;
            }
                
        }
        if (sum1 == 0) 
        {
            cout << "No" << endl;
        }
        else
        {
            if (n == x) 
            {
                if (sum1 % 2 == 0) 
                {
                    cout << "No" << endl;
                }
                else
                {
                    cout << "Yes" << endl;
                }
            }
            else 
            {
                if (x % 2 == 0 && sum2 == 0)
                {
                    cout << "No" << endl;
                }
                else 
                {
                    cout << "Yes" << endl;
                }
            }
        }
    }
    return 0;
}