time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an array aa consisting of nn integers a1,a2,…,ana1,a2,…,an. You want to split it into exactly kk 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 nn elements of the array aa must belong to exactly one of the kk subsegments.
Let's see some examples of dividing the array of length 55 into 33 subsegments (not necessarily with odd sums): [1,2,3,4,5][1,2,3,4,5] is the initial array, then all possible ways to divide it into 33 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];
- [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 kk 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 qq independent queries.
Input
The first line contains one integer qq (1≤q≤2⋅1051≤q≤2⋅105) — the number of queries. Then qq queries follow.
The first line of the query contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of elements in the array and the number of subsegments, respectively.
The second line of the query contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the ii-th element of aa.
It is guaranteed that the sum of nn over all queries does not exceed 2⋅1052⋅105 (∑n≤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 kk 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 kk integers r1r1, r2r2, ..., rkrk such that 1≤r1<r2<⋯<rk=n1≤r1<r2<⋯<rk=n, where rjrj is the right border of the jj-th segment (the index of the last element that belongs to the jj-th segment), so the array is divided into subsegments [1;r1],[r1+1;r2],[r2+1,r3],…,[rk−1+1,n][1;r1],[r1+1;r2],[r2+1,r3],…,[rk−1+1,n]. Note that rkrk is always nn but you should print it anyway.
Example
input
Copy
3 5 3 7 18 3 14 1 5 4 1 2 3 4 5 6 2 1 2 8 4 10 2
output
Copy
YES 1 3 5 NO NO
解题说明：题意是将长度为n的数组a划分成k个非空组，确保所有的组内之和都为奇数。做法是首先计算数组中奇数个数，如果奇数个数等于k，则每组一个奇数，满足需求。如果奇数个数比k大奇数个，则这些奇数一定会导致一个组的和变成偶数，不满足。如果奇数个数比k大偶数个，则满足要求。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include<iostream>
#include<algorithm>
#include <bits/stdc++.h>
using namespace std;
int f[200010], n, k, q, x;
int main()
{
cin >> q;
while (q--)
{
cin >> n >> k;
int cnt = 0,i;
for (i = 1; i <= n; i++)
{
cin >> x;
if (x & 1)
{
f[++cnt] = i;
}
}
if (cnt >= k && (cnt - k + 1) & 1)
{
printf("YES\n");
for ( i = 1; i < k; i++)
{
cout << f[i] << ' ';
}
cout << n << endl;
}
else
{
printf("NO\n");
}
}
return 0;
}