CADDi 2018 for Beginners题解

我这种zz大概也就只能写个beginner级的题解吧

C

类题

同样的预处理，然后枚举质因子，若该质因子p的次数大于等于n，就给答案乘p，次数-=n再判断，判断不成立在进行下一个

#include <bits/stdc++.h>
#define LL long long
#define int long long
#define db double
using namespace std;
 
const int MAXN = 500500;
const int MAXE = 400400;
const int INF = (1LL << 60 | 100);
 
template<typename T> inline void CheckMax(T &A, T B) {
	A < B ? A = B : A;
}
 
template<typename T> inline void CheckMin(T &A, T B) {
	A > B ? A = B : A;
}
 
template <typename T> inline void read(T &x) {
    int c = getchar();
    bool f = false;
    for (x = 0; !isdigit(c); c = getchar()) {
        if (c == '-') {
            f = true;
        }
    }
    for (; isdigit(c); c = getchar()) {
        x = x * 10 + c - '0';
    }
    if (f) {
        x = -x;
    }
}
 
LL n, m;
LL P[MAXN], cnt, num[MAXN];
 
void solve(LL x) {
	for(LL i = 2; i * i <= x; i++) {
		if(x % i == 0) {
			LL cur = 0;
			while(x % i == 0) {
				x /= i;
				cur++;
			}
			P[++cnt] = cur;
			num[cnt] = i;
		}
	}
	if(x > 1) P[++cnt] = 1;
}
 
signed main() {
	cin >> n >> m;
	if(n == 1) {
		cout << m << endl;
		return 0;
	}
	if(n > m) {
		cout << 1 << endl;
		return 0;
	}
	solve(m);
	LL ans = 1LL;
	for(int i = 1; i <= cnt; i++) {
		while(P[i] >= n) {
			ans = ans * num[i], P[i] -= n;
		}
	}
	cout << ans << endl;
	return 0;
}

D

蜜汁博弈论……全是偶数后手必胜……完了

#include <bits/stdc++.h>
#define LL long long
#define db double
using namespace std;
 
const int MAXN = 500500;
const int MAXE = 400400;
const int INF = 0x7FFFFFFF;
 
template<typename T> inline void CheckMax(T &A, T B) {
	A < B ? A = B : A;
}
 
template<typename T> inline void CheckMin(T &A, T B) {
	A > B ? A = B : A;
}
 
template <typename T> inline void read(T &x) {
    int c = getchar();
    bool f = false;
    for (x = 0; !isdigit(c); c = getchar()) {
        if (c == '-') {
            f = true;
        }
    }
    for (; isdigit(c); c = getchar()) {
        x = x * 10 + c - '0';
    }
    if (f) {
        x = -x;
    }
}

int n;
#define Fir puts("first")
#define Sec puts("second")

int a[MAXN], res; 
 
signed main() {
	read(n);
	for(int i = 1; i <= n; i++) {
		read(a[i]);
		if(a[i] & 1) res = 1;
	}
	if(res) Fir;
	else Sec;
	return 0;
}