CSP认证-现值计算、训练计划、JPEG 解码、聚集方差-Toy模板网

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CSP认证-现值计算、训练计划、JPEG 解码、聚集方差

T1现值计算

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
	int n;
	double rate;
	cin >> n >> rate;
	double t = 1.0, x, res = 0;
	for (int i = 1; i <= n + 1; ++i) {
		scanf("%lf", &x);
		res += t * x;
		t *= 1.0 / (1 + rate);
	}
	printf("%lf\n", res);
	return 0;
}

T2训练计划

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int p[N], t[N], level[N], idx[N], res1[N], res2[N];
bool is_leaf[N];
int main()
{
	freopen("a.in", "r", stdin);
	int n, m;
	scanf("%d%d", &n, &m);
	
	for (int i = 1; i <= m; ++i) scanf("%d", &p[i]), idx[i] = i, is_leaf[i] = true;
	for (int i = 1; i <= m; ++i) scanf("%d", &t[i]);
	// m^2
	for (int i = 1; i <= m; ++i) {
		int j = p[i];
		while (j) level[i]++, j = p[j];
	}
	// 拓扑 + 时间大到小
	sort(idx + 1, idx + 1 + m, [&](int i, int j) {
		return level[i] < level[j];
		});
	// 计算最早开始的时间，只需满足拓扑
	for (int i = 1; i <= m; ++i) {
		int j = idx[i];
		if (p[j]) res1[j] += (t[p[j]] + res1[p[j]]);
		else res1[j] = 1;
	}
	int ft = 0;
	for (int i = 1; i <= m; ++i) {
		printf("%d ", res1[i]);
		ft = max(ft, res1[i] + t[i] - 1);
	}
	printf("\n");
	if (ft > n) return 0;
	// 计算最晚开始时间
	else {
		memset(res2, 0x3f, sizeof res2);
		for (int i = m; i >= 1; --i) {
			int id = idx[i], pid = p[id];
			if (is_leaf[id]) res2[id] = n - t[id] + 1;
			if (pid) {
				is_leaf[pid] = false;
				res2[pid] = min(res2[id] - t[pid], res2[pid]);
			}
		}
		for (int i = 1; i <= m; ++i) {
			printf("%d ", res2[i]);
		}
		return 0;
	}

}

T3JPEG 解码

#include <bits/stdc++.h>
using namespace std;
#define PI acos(-1)
#define qrt sqrt(1.0 / 2.0)

int Q[10][10], mat[10][10];
double Q_2[10][10];
int n, T, row = 1, col = 1, cur = 0, nxt = 1;

void in_mat()
{
	for (int i = 1; i <= 8; ++i) {
		for (int j = 1; j <= 8; ++j) {
			scanf("%d", &Q[i][j]);
		}
	}
	scanf("%d%d", &n, &T);

	for (int i = 1; i <= n; ++i) {
		if (i < 36) {
			if (nxt % 2) scanf("%d", &mat[row--][col++]);
			else scanf("%d", &mat[row++][col--]);
			++cur;
			if (cur == nxt) {
				cur = 0;
				nxt++;
				if (nxt % 2) row = nxt, col = 1;
				else row = 1, col = nxt;
			}
		}
		else if (i == 36) {
			scanf("%d", &mat[row][col]);
			cur = 0;
			nxt = 7;
			row = 8, col = 2;
		}
		else {
			if (nxt % 2) scanf("%d", &mat[row--][col++]);
			else scanf("%d", &mat[row++][col--]);
			++cur;
			if (cur == nxt) {
				cur = 0;
				nxt--;
				if (nxt % 2) row = 8, col = 8 - nxt + 1;
				else row = 8 - nxt + 1, col = 8;
			}
		}

	}
}

void plus_mat()
{
	for (int i = 1; i <= 8; ++i) {
		for (int j = 1; j <= 8; ++j) {
			mat[i][j] *= Q[i][j];
		}
	}
}


void DCT(int m[10][10], double res[10][10])
{
	double au, av;
	for (int i = 1; i <= 8; ++i) {
		for (int j = 1; j <= 8; ++j) {
			double t = 0;
			for (int u = 0; u <= 7; ++u) {
				if (u == 0) au = qrt;
				else au = 1;
				for (int v = 0; v <= 7; ++v) {
					if (v == 0) av = qrt;
					else av = 1;
					t += 1.0 / 4.0 * au * av * m[u + 1][v + 1] * cos(PI / 8 * (i  - 1 + 1.0 / 2) * u)
						* cos(PI / 8 * (j  - 1 + 1.0 / 2) * v);
				}
			}
			res[i][j] = t;
		}
	}
	for (int i = 1; i <= 8; ++i) {
		for (int j = 1; j <= 8; ++j) {
			int t = round(res[i][j] + 128);
			if (t > 255) t = 255;
			else if (t < 0) t = 0;
			printf("%d ", t);
		}
		printf("\n");
	}
}

void out_mat()
{
	for (int i = 1; i <= 8; ++i) {
		for (int j = 1; j <= 8; ++j) {
			printf("%d ", mat[i][j]);
		}
		printf("\n");
	}
}



int main()
{
	freopen("a.in", "r", stdin);
	in_mat();

	if (T == 0) {
		out_mat();
	}
	else if (T == 1) {
		plus_mat();
		out_mat();
	}
	else {
		plus_mat();
		DCT(mat, Q_2);
	}
	return 0;
}

T4聚集方差

65分解法如下：文章来源地址https://www.toymoban.com/news/detail-518501.html

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 300100;
int h[N], e[N], ne[N], a[N], idx;
LL ans[N];
void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

// 合并两个有序数组O(n)
void merge(vector<int>& dest, vector<int>& v)
{
	vector<int> v3(dest.size() + v.size());
	if (dest.empty()) dest = v;
	else {
		int p1 = 0, p2 = 0, i = 0;
		while (p1 != dest.size() && p2 != v.size()) {
			if (dest[p1] < v[p2]) v3[i++] = dest[p1++];
			else v3[i++] = v[p2++];
		}
		if (p1 == dest.size()) {
			while(p2 != v.size())
				v3[i++] = v[p2++];
		}
		else{
			while (p1 != dest.size())
				v3[i++] = dest[p1++];
		}
	}
	dest = v3;
}

LL calc(vector<int>& v)
{
	int n = v.size();
	if (n == 1) return LL(0);
	LL res = 0;
	for (int i = 0; i < n; ++i) {
		int min_dis = 2e9;
		if (i - 1 >= 0) 
			min_dis = abs(v[i - 1] - v[i]);
		if (i + 1 < n) 
			min_dis = min(min_dis, abs(v[i + 1]) - v[i]);
		res += (LL)min_dis * min_dis;
	}
	return res;
}

vector<int> dfs(int u)
{
	vector<int> res;
	res.push_back(a[u]);
	for (int i = h[u]; ~i; i = ne[i]) {
		vector<int> t = dfs(e[i]);
		merge(res, t);
	}
	ans[u] = calc(res);
	return res;
}


int main()
{
	int n, pp;
	scanf("%d", &n);
	memset(h, -1, sizeof h);
	for (int i = 2; i <= n; ++i) {
		scanf("%d", &pp);
		add(pp, i);
	}
	for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	dfs(1);
	for (int i = 1; i <= n; ++i) {
		printf("%lld\n", ans[i]);
	}
	return 0;
}