6.18总结 - 技术栈

省赛排位赛2：

思路：

设两个方程直接解出来就行

代码：

复制代码

#include<bits/stdc++.h>
using namespace std;
int n, m; 
int main()
{
    int n, m;
    int ans1, ans2;
    cin >> n >> m;
    ans1 = n - (-3 + sqrt(3 * 3 + 4 * 2 * (n + m))) / 2;
    ans2 = n - (-3 - sqrt(3 * 3 + 4 * 2 * (n + m))) / 2;
    if (ans1 <= n && ans2 >= 0) cout << ans1;
    else cout << ans2;
}

省赛排名赛2 - Virtual Judge

代码：

复制代码

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
ll n;
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		cin >> n;
		ll ans = 0;
		ll nn = n;
		ll i = 1;
		while (nn)
		{
			ans += n / i;
			i <<= 1ll;
			nn >>= 1ll;
		}
		cout << ans << endl;
	}
	return 0;
}

省赛排名赛2 - Virtual Judge

代码：

复制代码

#include<bits/stdc++.h>
using namespace std;
int a[10000005];
int main()
{
	int n, p;
	cin >> n >> p;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		a[i] += a[i - 1];
	}
	int ans = 0;
	for (int i = 1; i < n; i++)
	{
		ans = max(ans, (a[i] % p + (a[n] - a[i]) % p));
	}
	cout << ans << endl;
	return 0;
}

省赛排名赛2 - Virtual Judge

思路：

需要得到每个(x,y)下的最大公约数，再找出一定的规律

代码：

复制代码

#include<bits/stdc++.h>
using namespace std;
long long t, a, b, c, n, ca, cb, cc;
int gcd(int x, int y)
{
	if (x < y)swap(x, y);
	if (x % y)return gcd(y, x % y);
	return y;
}
int main()
{
	cin >> t;
	while (t--)
	{
		cin >> n >> a >> b;
		c = a * b / gcd(a, b);
		ca = n / a, cb = n / b, cc = n / c;
		ca -= cc, cb -= cc;
		cout << (n + n - ca + 1) * ca / 2 - (1 + cb) * cb / 2 << endl;
	}
	return 0;
}

省赛排名赛2 - Virtual Judge

代码：

复制代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[1000010], n;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin>>a[i], a[i] ^= a[i - 1];
	ll ans = 0;
	for (int i = 0, x = (1 << i); i < 32; i++, x <<= 1)
	{
		int cnt = 0;
		for (int j = 1; j <= n; j++)
			if ((a[j] >> i) & 1) cnt++;
		ans += 1LL * cnt * (n + 1 - cnt) * x;
	}
	cout << ans << endl;
	return 0;
}

省赛排名赛2 - Virtual Judge

思路：

给定一个能达到数据范围的斐波拉契数组，包含所有数据范围内的斐波拉契数，再进行几个特判

代码：

复制代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll f[100];
string check(ll n, ll x, ll y)
{
    if (x == 1 && y == 1)
    {
        return "YES";
    }
    if (y <= f[n] && y > f[n - 1])
    {
        return "NO";
    }
    if (y > f[n])
    {
        y = y - f[n];
    }
    return check(n - 1, y, x);
}
int main()
{
    ll t, n, x, y;
    f[0] = 1;
    f[1] = 1;
    for (int i = 2; i <= 45; i++)
        f[i] = f[i - 1] + f[i - 2];
    cin >> t;
    while (t--)
    {
        cin >> n >> x >> y;
        cout << check(n, x, y) << endl;;
    }
    return 0;
}

省赛排名赛2 - Virtual Judge

思路：

01背包问题，就题目改了一下，直接用模板就行

代码：

复制代码

#include<bits/stdc++.h>
using namespace std;
int h, t, n, a[401], z[401], l[501], dp[401][401];
int main() {
    cin >> h >> t >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i] >> z[i] >> l[i];
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = h; j >= a[i]; j--)
        {
            for (int k = t; k >= z[i]; k--)
            {
                dp[j][k] = max(dp[j][k], dp[j - a[i]][k - z[i]] + l[i]);
            }
        }
    }
    cout << dp[h][t];
    return 0;
}