徐州网站开发案例,门户网站免费奖励自己,网站开发人员招聘,中国建设银行笔试确认网站cf1556B B. Take Your Places!
题意#xff1a;
有n个数#xff0c;你可以将相邻两个数交换#xff0c;使得奇偶性一样的数不相邻。问最少操作步数
题解#xff1a;
最终排列无非是#xff1a;奇#xff0c;偶#xff0c;奇…或者偶#xff0c;奇#xff0c;偶… …cf1556B B. Take Your Places!
题意
有n个数你可以将相邻两个数交换使得奇偶性一样的数不相邻。问最少操作步数
题解
最终排列无非是奇偶奇…或者偶奇偶… 如果奇数偶数我们就按照第一个排如果偶数奇数我们就按照第二个排。如果两者相等两种排列都测一遍取较小值
代码
// Problem: B. Take Your Places!
// Contest: Codeforces - Deltix Round, Summer 2021 (open for everyone, rated, Div. 1 Div. 2)
// URL: https://codeforces.com/contest/1556/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// Data:2021-08-31 23:46:12
// By Jozky#include bits/stdc.h
#include unordered_map
#define debug(a, b) printf(%s %d\n, a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pairint, int PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll 1e18;
const int INF_int 0x3f3f3f3f;
void read(){};
template typename _Tp, typename... _Tps void read(_Tp x, _Tps... Ar)
{x 0;char c getchar();bool flag 0;while (c 0 || c 9)flag| (c -), c getchar();while (c 0 c 9)x (x 3) (x 1) (c ^ 48), c getchar();if (flag)x -x;read(Ar...);
}
template typename T inline void write(T x)
{if (x 0) {x ~(x - 1);putchar(-);}if (x 9)write(x / 10);putchar(x % 10 0);
}
void rd_test()
{
#ifdef LOCALstartTime clock();freopen(in.txt, r, stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime clock();printf(\nRun Time:%lfs\n, (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn 1e5 9;
int a[maxn];
int b[maxn];
int main()
{//rd_test();int t;read(t);while (t--) {int n;read(n);vectorint q1; //奇数vectorint q2; //偶数for (int i 0; i n; i) {read(a[i]);if (a[i] % 2 1) //奇数q1.push_back(i);elseq2.push_back(i);}if (n 1)printf(0\n);else if (abs((int)q1.size() - (int)q2.size()) 1)printf(-1\n);else if ((int)q1.size() (int)q2.size()) {int pos 0;ll sum 0;for (int i 0; i q1.size(); i) {sum abs(q1[i] - pos);pos 2;}printf(%lld\n, sum);}else if ((int)q1.size() (int)q2.size()) {int pos 0;ll sum 0;for (int i 0; i q2.size(); i) {sum abs(q2[i] - pos);pos 2;}printf(%lld\n, sum);}else if ((int)q1.size() (int)q2.size()) {int pos 0;ll sum 0;for (int i 0; i q1.size(); i) {sum abs(q1[i] - pos);pos 2;}pos 0;ll sum2 0;for (int i 0; i q2.size(); i) {sum2 abs(q2[i] - pos);pos 2;}printf(%lld\n, min(sum, sum2));}}return 0;//Time_test();
}
