bzoj 2751 [HAOI2012]容易题(easy) 数学

题面

题目传送门

解法

考虑当所有数都没有限制的时候，答案为\(\sum\prod_{i=1}^na_i\)

考虑将式子拆开并且运用乘法分配律，答案显然为\((\frac{n(n+1)}{2})^m\)

有限制类似

注意去重

代码

#include <bits/stdc++.h>
#define Mod 1000000007
#define int long long
#define N 100010
using namespace std;
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
	int x, y;
	bool operator < (const Node &a) const {
		if (x == a.x) return y < a.y;
		return x < a.x;
	}
} a[N];
int Pow(int x, int y) {
	int ret = 1;
	while (y) {
		if (y & 1) ret = ret * x % Mod;
		y >>= 1, x = x * x % Mod;
	}
	return ret;
}
main() {
	int n, m, k; read(n), read(m), read(k);
	for (int i = 1; i <= k; i++)
		read(a[i].x), read(a[i].y);
	sort(a + 1, a + k + 1);
	int tot = 0, ans = 1, s = 1, sum = n * (n + 1) / 2 % Mod;
	map <int, int> h;
	for (int i = 1; i <= k; i++) {
		if (a[i].x != a[i - 1].x)
			tot++, ans = ans * s % Mod, s = sum, h.clear();
		if (!h.count(a[i].y)) s = (s - a[i].y + Mod) % Mod, h[a[i].y] = 1;
	}
	ans = ans * s % Mod * Pow(sum, m - tot) % Mod;
	cout << ans << "\n";
	return 0;
}