One-by-one to solve the Sequence Problem (2)

最新推荐文章于 2025-01-23 23:46:20 发布

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最新推荐文章于 2025-01-23 23:46:20 发布

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分类专栏： algorithm-design

本文链接：https://blog.youkuaiyun.com/baidu_17640849/article/details/68925716

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本文探讨了一个涉及整数数组的问题，即通过删除相邻且和为奇数的元素来尽可能缩短序列，目标是最小化最终序列的长度。文章提供了一个具体实例，并给出了一段C++代码示例，用于解决这个问题。

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#1399 : Shortening Sequence (from hihocoder) Stack

时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

There is an integer array A1, A2 …AN. Each round you may choose two adjacent integers. If their sum is an odd number, the two adjacent integers can be deleted.

Can you work out the minimum length of the final array after elaborate deletions?

输入

The first line contains one integer N, indicating the length of the initial array.

The second line contains N integers, indicating A1, A2 …AN.

For 30% of the data：1 ≤ N ≤ 10

For 60% of the data：1 ≤ N ≤ 1000

For 100% of the data：1 ≤ N ≤ 1000000, 0 ≤ Ai ≤ 1000000000

输出

One line with an integer indicating the minimum length of the final array.

样例提示

(1,2) (3,4) (4,5) are deleted.

样例输入

7
1 1 2 3 4 4 5

样例输出

Code Snippet

#define LOG(x) cout << #x << " = " << (x) << endl
#define PRINTLN(x) cout << (x) << endl
#define MEM(x, y) memset((x), (y), sizeof((x)))
#include <bits/stdc++.h>
using namespace std;
const double PI = 2*acos(0);
typedef long long ll;
typedef complex<double> Complex;
int nextInt()
{
    int x;
    scanf("%d", &x);
    return x;
}
ll nextLL()
{
    ll x;
    scanf("%lld", &x);
    return x;
}
//TEMPLATE

//MAIN
int main()
{
    //freopen("in.txt", "r", stdin);
    int n;
    while (scanf("%d", &n) != EOF) {
         stack<ll> s;
    int ans = 0;
    for (int i = 0; i < n; i++) {
      ll t = nextLL();
      if (!s.empty() && (s.top() + t) % 2) {
            s.pop();
            ans += 2;
          } else s.push(t);
        }
        PRINTLN(n - ans);
    }
}