Codeforces Round #680 (Div. 2, based on Moscow Team Olympiad) A ~ C

A

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define fi firsqt
#define se second
#define inf 0x3f3f3f3f
const int N = 55;
int a[N], b[N];
 
int main() {	
	IO;
	int _;
	cin >> _;
	while (_--) {
		int n, x;
		cin >> n >> x;	
		for (int i = 0; i < n; ++i) cin >> a[i];
		for (int i = 0; i < n; ++i) cin >> b[i];
		int flag = 1;
		for (int i = 0; i < n; ++i) 
			if (a[i] + b[n - 1 - i]	> x) {
				flag = 0;
				break;
			}
		if (flag) puts("YES");
		else puts("NO");
	}
	return 0;
}

B

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define fi firsqt
#define se second
#define inf 0x3f3f3f3f
const int N = 55;
int a[N], b[N];
 
int main() {	
	IO;
	int _;
	cin >> _;
	while (_--) {
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		int ans = max(a + b, c + d);
		cout << ans << endl;
	}
	return 0;
}

C

参考于 : https://www.cnblogs.com/zwjzwj/p/13912656.html

题意

给出两个数p, q, 找到最大的x, 使得x是p的因子且x不是q的倍数.

思路

分两种情况 :

1. : 如果q不是p的因子, 答案就是p.

2. : 当q是p的因子时, 去枚举q的质因子, 得到该质因子在q中最高次, 少去一次后乘上p将该质因子除干净得到的数. 例如 : 对于质因子i, 设i在p中最高有a次, 在q中最高有b次.

\[x = \frac{p}{i^a} * i^{b-1} = p * i^{b - 1 - a} \]

注 : 因为p是q的倍数所以必定满足a >= b.

代码

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define fi firsqt
#define se second
#define inf 0x3f3f3f3f


ll qpow(ll a, ll b) {
	ll res = 1;
	while (b) {
		if (b & 1) res = res * a;
		b >>= 1;
		a = a * a;
	}
	return res;
}
		

int main() {	
	IO;
	int _;
	cin >> _;
	while (_--) {
		ll p, q;
		cin >> p >> q;
		ll ans = 0;
		if (p % q != 0) ans = p;
		else {
			for (int i = 2; i <= q / i; ++i) {
				if (q % i == 0) {
					int cnt = 0;
					while (q % i == 0) q /= i, ++cnt;
					ll x = p;
					while (x % i == 0) x /= i;
					ans = max(ans, x * qpow(i, cnt - 1));
				}
			}
			if (q > 1) {
				ll x = p;
				while (x % q == 0) x /= q;
				ans = max(ans, x);
			}
		}
		cout << ans << endl;	
	}
	return 0;
}