B. Odd Sum Segments
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Example
inputCopy
3
5 3
7 18 3 14 1
5 4
1 2 3 4 5
6 2
1 2 8 4 10 2
outputCopy
YES
1 3 5
NO
NO
题意：
给你一个n个数的数组，让你分成k个部分，使每一部分的sum和是奇数
思路：
容易知道，想让sum和为奇数，这么这部分一定有奇数个奇数。
所以想构造成k个部分的条件是 if((sum-k)%2==0) ( sum是奇数的个数)
然后从后开始贪心的分成ｋ个部分即可，
本题坑点：要求最后一个ｒ一定是 n 这里wa了好几次。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int a[maxn];
int n,k;
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
int t;
gg(t);
while(t--)
{
gg(n);gg(k);
int sum=0;
repd(i,1,n)
{
gg(a[i]);
a[i]%=2;
sum+=a[i];
}
if(sum<k)
{
printf("NO\n");
continue;
}
if((sum-k)%2!=0)
{
printf("NO\n");
continue;
}
printf("YES\n");
std::vector<int> ans;
for(int i=n;i>=1;i--)
{
if(k)
{
if(a[i])
{
ans.push_back(i);
// printf("
// %d ",i);
k--;
}
}
}
ans[0]=n;
for(int i=sz(ans)-1;i>=0;--i)
{
printf("%d ",ans[i] );
}
printf("\n");
// cout<<endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}原文地址：https://www.cnblogs.com/qieqiemin/p/11247855.html