Codeforces 1196B Odd Sum Segments

题意：给一串序列，需要将序列划分成k组，要求每组连续的子序和为奇数，存在答案则输出答案每个区间的右端点
思路：还是先想到暴力回溯，结果还是超时了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
const int inf = 0x3f3f3f3f;
ll n, q, k, a[maxn], ok, sum[maxn];
vector<int> res;
void solve(int i, int cnt)
{
	if (ok)
		return;
	if (cnt == k && res[k-1] == n) {
		ok = 1;
		printf("YES\n");
		for (int t = 0; t < res.size(); t++)
			printf("%d ", res[t]);
		printf("\n");
		return;
	}
	for (int t = i+1; t <= n; t++) {
		if ((sum[t]-sum[i])&1) {
			res.push_back(t);
			solve(t, cnt+1);
			res.pop_back();
		}
		if (n - t < k - cnt) {
			return;
		}
	}
}
int main()
{
	while (cin >> q) {
		for (int i = 0; i < q; i++) {
			cin >> n >> k;
			sum[0] = 0;
			for (int j = 1; j <= n; j++) {
				cin >> a[j];
				sum[j] = sum[j-1] + a[j];
			}
			ok = 0;
			solve(0, 0);
			if (!ok)
				cout << "NO" << endl;
		}
	}	
	return 0;
}

看了别人的答案，发现只要统计奇数的个数，只要奇数个数与k同奇或同偶即可。
简单证明：若k为2，奇数个数为奇数，若取其中一个区间和为奇数，则另一个区间必然为偶数不成立；若奇数个数为偶数，则两区间都满足区间和为奇数成立。若k为3，奇数个数为奇数，则取一个区间后剩下的区间和为偶数，从里面可以再取出两个区间和为奇数的区间，故成立。
由于区间和为奇数，右端点只需要输出奇数就行

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
const int inf = 0x3f3f3f3f;
ll n, q, k, a[maxn], ok, sum[maxn];
int main()
{
	while (cin >> q) {
		for (int i = 0; i < q; i++) {
			vector<int> res;
			cin >> n >> k;
			sum[0] = 0;
			for (int j = 1; j <= n; j++) {
				cin >> a[j];
			}
			for (int j = 1; j <= n; j++)
				if (a[j]&1)
					res.push_back(j);
			int t = res.size();
			if (res.size() < k || (res.size()&1) != (k&1))
				cout << "NO" << endl;
			else {
				printf("YES\n");
				for (int j = 0; j < k-1; j++)
					printf("%d ", res[j]);
				printf("%d\n", n);
			}
		}
	}	
	return 0;
}