CodeForces - 1196B. Odd Sum Segments(思维/基础数论)

You are given an array a consisting of n integers a1,a2,…,an. You want to split it into exactly k non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to rearrange (shuffle) the elements of a given array. Each of the n elements of the array a must belong to exactly one of the k subsegments.

Let’s see some examples of dividing the array of length 5 into 3 subsegments (not necessarily with odd sums): [1,2,3,4,5] is the initial array, then all possible ways to divide it into 3 non-empty non-intersecting subsegments are described below:

[1],[2],[3,4,5];
[1],[2,3],[4,5];
[1],[2,3,4],[5];
[1,2],[3],[4,5];
[1,2],[3,4],[5];
[1,2,3],[4],[5].
Of course, it can be impossible to divide the initial array into exactly k subsegments in such a way that each of them will have odd sum of elements. In this case print “NO”. Otherwise, print “YES” and any possible division of the array. See the output format for the detailed explanation.

You have to answer q independent queries.

Input
The first line contains one integer q (1≤q≤2⋅105) — the number of queries. Then q queries follow.

The first line of the query contains two integers n and k (1≤k≤n≤2⋅105) — the number of elements in the array and the number of subsegments, respectively.

The second line of the query contains n integers a1,a2,…,an (1≤ai≤109), where ai is the i-th element of a.

It is guaranteed that the sum of n over all queries does not exceed 2⋅105 (∑n≤2⋅105).

Output
For each query, print the answer to it. If it is impossible to divide the initial array into exactly k subsegments in such a way that each of them will have odd sum of elements, print “NO” in the first line. Otherwise, print “YES” in the first line and any possible division of the array in the second line. The division can be represented as k integers r1, r2, …, rk such that 1≤r1<r2<⋯<rk=n, where rj is the right border of the j-th segment (the index of the last element that belongs to the j-th segment), so the array is divided into subsegments [1;r1],[r1+1;r2],[r2+1,r3],…,[rk−1+1,n]. Note that rk is always n but you should print it anyway.

Examples input
3
5 3
7 18 3 14 1
5 4
1 2 3 4 5
6 2
1 2 8 4 10 2
Examples output
YES
1 3 5
NO
NO

题意:

给你你个数, 分成k块, 求是否可以每一块的和都为奇数

题解:

对于奇偶的问题, 要首先想到奇偶的交换律, 即
奇+奇 = 偶
偶+偶 = 偶
奇+偶 = 奇
这样的话, 我们考虑, 在奇数的个数>=k时, 当且仅当奇数的个数%2同k%2相同时, 即说明一定有解

代码:

#include <bits/stdc++.h>
#define LL long long
const int N = 500000 + 5;
using namespace std;

LL a[N];
int main()
{
    int q;
    scanf("%d", &q);
    while (q--)
    {
        LL n, k;
        scanf("%lld%lld", &n, &k);

        LL sum = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%lld", &a[i]);
            if (a[i] % 2)
                sum++;
        }

        if (sum < k)
            printf("NO\n");
        else
        {
            if ((sum - (k + 1)) % 2 == 0)
                printf("NO\n");
            else
            {
                printf("YES\n");
                if (k > 1)
                {
                    int cnt = 0;
                    LL sum = 0;
                    for (int i = 1; i <= n; i++)
                    {
                        sum += a[i];
                        if (sum % 2 == 1)
                        {
                            cnt++;
                            printf("%d ", i);
                            sum = 0;
                        }
                        if (cnt >= k - 1)
                            break;
                    }
                }
                printf("%d\n", n);
            }
        }
    }
    return 0;
}