Codeforces Round 998 (Div. 3)(部分题解)

补题链接

A. Fibonacciness

思路：了解清楚题意，求得是最大的斐波那契的度，数组只有5个数(最多度为3)，能列出其对应的式子

 或 或

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    int n, m, k;
    vector<int> a(4);
    set<int> s;
    for (int i= 0; i<4; i++) cin >> a[i];
    s.insert(a[0]+a[1]);
    s.insert(a[2]-a[1]);
    s.insert(a[3]-a[2]);
    cout << 4-s.size() << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    
    return 0;
}

B. Farmer John's Card Game

 思路：

我的思路偏向于暴力一点，先排好序，并记录位置，再遍历跑一边n*m检查有没有不符合条件的。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define vvi vector<vi>

void solve()
{
	int n, m, k;
	cin >> n >> m;
	vvi a(n+1, vi(m+1));
	map<int, int> mp;
	for (int i = 1; i <= n; ++i){
		for (int j = 1; j <= m; ++j){
			cin >> a[i][j];
		}
		sort(a[i].begin()+1, a[i].end());
		mp[a[i][1]] = i;
	}
	sort(a.begin()+1, a.end(), [](vi &x, vi &y){
		return x[1] < y[1];
	});
	int flag = 0;
	int last = -1;
	for (int i = 1; i <= m; ++i){
		for (int j = 1; j <= n; ++j){
			if (a[j][i] < last) {
				flag = 1;
				break;
			}
			else last = a[j][i];
		}
		if (flag) break;
	}
	if (flag) cout << -1 << endl;
	else {
		for (int i = 1; i <= n; ++i){
			cout << mp[a[i][1]] << " \n"[i == n];
		}
	}
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	int t = 1;
	cin >> t;
	while (t--)
	{
		solve();
	}
	
	return 0;
}

题解的思路会巧一点不用重复遍历。

附上关键代码：

for (vll &we : ve) {
        for (ll &i : we) cin >> i;
        ll minN = *min_element(we.begin(), we.end());
        if (minN < n) p[minN] = c++;
        val &= minN < n;
        sort(we.begin(), we.end());
        ll last = we[0]-n;
        for (ll i : we) {
            val &= last+n == i;
            last = i;
        }
    }

C. Game of Mathletes

思路：

记录数字出现的次数，枚举k的一半，（k为偶数特判一下）取mp[i],mp[k-i]的较小值。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>

void solve()
{
	int n, m, k;
	cin >> n >> k;
	vi a(n);
	map<int, int> mp;
	for (int i = 0; i < n; i++){
		cin >> a[i];
		mp[a[i]]++;
	}
	int ans = 0;
	for (int i = 1; i<=k/2; i++){
		if (k % 2 ==0 && i == k/2) ans += mp[i]/2;
		else ans += min(mp[i], mp[k-i]);
		
	}
	cout << ans << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	int t = 1;
	cin >> t;
	while (t--)
	{
		solve();
	}
	
	return 0;
}

D. Subtract Min Sort

思路：

从头开始遍历，如果a[i-1]>a[i]就不成立，

否则a[i]和a[i-1]都同时减去min(a[i-1],a[i])

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define endl '\n'
void solve()
{
	int n, m, k;
	cin >> n;
	vi a(n);
	int flag = 0;
	for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 1; i<n; i++){
		if (a[i-1] > a[i]) {
			flag = 1;
			break;
		}
		else {
			int mi = min(a[i-1], a[i]);
			a[i-1] -= mi;
			a[i]-=mi;
		}
	}
	if (flag) cout << "NO" << endl;
	else cout << "YES" << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	int t = 1;
	cin >> t;
	while (t--)
	{
		solve();
	}
	
	return 0;
}