BestCoder Round #8

题目链接

A：签到题不多说

B：矩阵快速幂，奇数项的式子为f(n) = 4 * f(n - 1) + 1，偶数项是奇数项的两倍，然后构造矩阵为4 1 0 1进行快速幂即可

C：dp+树状数组加速，dp[i][j]表示以i为结尾长度为j的种数，然后把数字离散化掉，每次状态转移都需要从前一个区间和转移过来，所以可以利用树状数组维护

代码：

A：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;

typedef long long ll;
ll num[105];
set<ll> save;

int n;
int main() {
	while (~scanf("%d", &n)) {
		save.clear();
		for (int i = 0; i < n; i++) {
			scanf("%I64d", &num[i]);
			for (int j = 0; j < i; j++) {
				save.insert(num[i] + num[j]);
			}
		}
		ll ans = 0;
		for (set<ll>::iterator it = save.begin(); it != save.end(); it++) {
			ans += *it;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}

B：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;

typedef long long ll;

ll n, m;

struct mat {
	ll v[2][2];
	mat() {memset(v, 0, sizeof(v));}
	mat operator * (mat c) {
		mat ans;
		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < 2; j++) {
				for (int k = 0; k < 2; k++) {
					ans.v[i][j] = (ans.v[i][j] + v[i][k] * c.v[k][j]) % m;
				}
			}
		}
		return ans;
	}
};

mat pow_mod(mat A, ll k) {
	mat ans;
	for (int i = 0; i < 2; i++) ans.v[i][i] = 1;
	while (k) {
		if (k&1) ans = ans * A;
		A = A * A;
		k >>= 1;
	}
	return ans;
}

int main() {
	while (cin >> n >> m) {
		mat A;
		A.v[0][0] = 4; A.v[0][1] = 1;
		A.v[1][0] = 0; A.v[1][1] = 1;
		A = pow_mod(A, (n + 1) / 2);
		ll ans = A.v[0][1];
		ans %= m;
		if (n % 2 == 0) ans = ans * 2 % m;
		cout << ans << endl;
	}
	return 0;
}

C：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define lowbit(x) (x&(-x))

const int N = 10005;
const int M = 105;
const int MOD = 123456789;

int n, m, dp[N][M], bit[105][N];

void add(int *bit, int x, int v) {
	while (x < N) {
		bit[x] = (bit[x] + v) % MOD;
		x += lowbit(x);
	}
}

int get(int *bit, int x) {
	int ans = 0;
	while (x) {
		ans = (ans + bit[x]) % MOD;
		x -= lowbit(x);
	}
	return ans;
}

struct Num {
	int val, rank, id;
} num[N];

bool cmpid(Num a, Num b) {
	return a.id < b.id;
}

bool cmpval(Num a, Num b) {
	return a.val < b.val;
}

int main() {
	while (~scanf("%d%d", &n, &m)) {
		memset(bit, 0, sizeof(bit));
		for (int i = 1; i <= n; i++) {
			scanf("%d", &num[i].val);
			num[i].id = i;
		}
		sort(num + 1, num + n + 1, cmpval);
		num[1].rank = 2;
		for (int i = 2; i <= n; i++) {
			num[i].rank = num[i - 1].rank;
			if (num[i].val > num[i - 1].val)
				num[i].rank++;
		}
		sort(num + 1, num + n + 1, cmpid);
		add(bit[0], 1, 1);
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= n; i++) {
			for (int j = min(m, i); j >= 1; j--) {
				int tmp = get(bit[j - 1], num[i].rank - 1);
				dp[i][j] = (dp[i][j] + tmp) % MOD;
				add(bit[j], num[i].rank, tmp);
			}
		}
		int ans = 0;
		for (int i = 1; i <= n; i++)
			ans = (ans + dp[i][m]) % MOD;
		printf("%d\n", ans);
	}
	return 0;
}