Codeforces A. Sign Flipping (思维 / 构造) (Global Round 9)

传送门

题意： 有一初始数组a，你可改变其中元素的符合，使得其至少有一半的元素满足a[i] < a[i + 1]，至少有一半满足a[i] > a[i + 1]。最后数组构造的数组。

思路： 只要一正一负推移下去就一定满足条件。

代码实现：

#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n;
int a[N];

signed main()
{
    IOS;

    cin >> t;
    while(t --){
        cin >> n;
        for(int i = 0; i < n; i ++){
            cin >> a[i];
            if(i % 2 && a[i] < 0) a[i] = -a[i];
            if(i % 2 == 0 && a[i] > 0) a[i] = -a[i];
        }
        for(int i = 0; i < n; i ++)
            cout << a[i] << " ";
        cout << endl;
    }

    return 0;
}

本文地址：https://blog.csdn.net/Satur9/article/details/107215216