第十六届蓝桥杯大赛软件赛省赛 C++ 大学 B 组部分题解

赛时参加的是Python组，这是赛后写的题解，还有两题暂时还不会，待更新

题目链接题目列表 - 洛谷 | 计算机科学教育新生态

A 移动距离

答案：1576

C 可分解的正整数

Python3

import itertools
from functools import cmp_to_key

import sys

input = sys.stdin.readline
sys.set_int_max_str_digits(100000)


def I():
    return input().strip()


def II():
    return int(I())


def MII():
    return map(int, I().split())


def LMII():
    return list(MII())


def solve():
    n = II()
    data = LMII()
    res = 0
    for x in data:
        if x != 1:
            res += 1
    print(res)


T = 1
for _ in range(T):
    solve()

C++

#include"iostream"
using namespace std;

int main() {
	int n = 0, x, res = 0;
	cin >> n;
	for (int i = 0;i < n;i++) {
		cin >> x;
		if (x != 1)res++;
	}
	cout << res;
}

D 产值调整

Python3

import itertools
from functools import cmp_to_key

import sys

input = sys.stdin.readline
sys.set_int_max_str_digits(100000)


def I():
    return input().strip()


def II():
    return int(I())


def MII():
    return map(int, I().split())


def LMII():
    return list(MII())


def solve():
    t = II()
    for i in range(t):
        a, b, c, k = MII()
        for j in range(k):
            if a == b == c:
                break
            a, b, c = (b + c) // 2, (a + c) // 2, (a + b) // 2
        print(a, b, c)


T = 1
for _ in range(T):
    solve()

C++

#include"iostream"
using namespace std;

int main() {
	int t;
	cin >> t;
	for (int i = 0;i < t;i++) {
		int a, b, c, k;
		cin >> a >> b >> c >> k;
		for (int j = 0;j < k;j++) {
			if (a == b && b == c)break;
			int aa = (b + c) / 2;
			int bb = (a + c) / 2;
			int cc = (a + b) / 2;
			a = aa;
			b = bb;
			c = cc;
		}
		cout << a << " " << b << " " << c << endl;
	}
}

E 画展布置

Python3

import itertools
from functools import cmp_to_key

import sys

input = sys.stdin.readline
sys.set_int_max_str_digits(100000)


def I():
    return input().strip()


def II():
    return int(I())


def MII():
    return map(int, I().split())


def LMII():
    return list(MII())


def solve():
    n, m = MII()
    data = LMII()
    data.sort()
    tmp = [0] * (n - 1)
    for i in range(n - 1):
        tmp[i] = data[i + 1] ** 2 - data[i] ** 2
    res = 0
    for i in range(m - 1):
        res += tmp[i]
    idx = 0
    now = res
    for i in range(m - 1, n - 1):
        now = now + tmp[i] - tmp[idx]
        res = min(res, now)
        idx += 1
    print(res)


T = 1
for _ in range(T):
    solve()

C++

#include"iostream"
#include"math.h"
#include"algorithm"
#include"vector"
using namespace std;

int main() {
	int n, m;
	cin >> n >> m;
	vector<int> data;
	for (int i = 0;i < n;i++) {
		int x;
		cin >> x;
		data.push_back(x);
	}
	sort(data.begin(), data.end());
	int tmp[100005] = { 0 };
	for (int i = 0;i < n - 1;i++)tmp[i] = pow(data[i + 1], 2) - pow(data[i], 2);
	long long res = 0;
	for (int i = 0;i < m - 1;i++)res += tmp[i];
	int idx = 0;
	long long now = res;
	for (int i = m - 1;i < n - 1;i++) {
		now = now + tmp[i] - tmp[idx];
		res = min(res, now);
		idx++;
	}
	cout << res << endl;
}

F 水质检测

Python3

import itertools
from functools import cmp_to_key

import sys

input = sys.stdin.readline
sys.set_int_max_str_digits(100000)


def I():
    return input().strip()


def II():
    return int(I())


def MII():
    return map(int, I().split())


def LMII():
    return list(MII())


def solve():
    s1 = I()
    s2 = I()
    st = -1
    en = 0
    for i in range(len(s1)):
        if s1[i] == "#" or s2[i] == "#":
            if st == -1:
                st = i
            else:
                en = i
    # print(st, en)
    if st == -1 or en == 0:
        print(0)
    else:
        res = 0
        tmp = 0
        for i in range(st, en + 1):
            # print(tmp,s1[i],s2[i])
            if s1[i] == s2[i] == '.':
                res += 1
            elif s1[i] == '.':
                if tmp == 1:
                    res += 1
                    tmp = 0
                else:
                    tmp = 2
            elif s2[i] == ".":
                if tmp == 2:
                    res += 1
                    tmp = 0
                else:
                    tmp = 1
            else:
                tmp = 0
            # print(i, res)
        print(res)


T = 1
for _ in range(T):
    solve()

C++

#include"iostream"
#include"math.h"
#include"algorithm"
#include"vector"
#include"string"
using namespace std;

int main() {
	string s1;
	string s2;
	cin >> s1;
	cin >> s2;
	int st = -1;
	int en = 0;
	for (int i = 0;i < s1.size();i++) {
		if (s1[i] == '#' || s2[i] == '#') {
			if (st == -1)st = i;
			else en = i;
		}
	}
	if (st == -1 || en == 0)cout << 0;
	else {
		int res = 0;
		int tmp = 0;
		for (int i = st;i < en + 1;i++) {
			if (s1[i] == '.' && s2[i] == '.') {
				res++;
			}
			else if (s1[i] == '.') {
				if (tmp == 1) {
					res++;
					tmp = 0;
				}
				else tmp = 2;
			}
			else if (s2[i] == '.') {
				if (tmp == 2) {
					res++;
					tmp = 0;
				}
				else tmp = 1;
			}
			else tmp = 0;
		}
		cout << res;
	}
}

G 生产车间

Python3

import itertools
from functools import cmp_to_key

import sys

input = sys.stdin.readline
sys.set_int_max_str_digits(100000)


def I():
    return input().strip()


def II():
    return int(I())


def MII():
    return map(int, I().split())


def LMII():
    return list(MII())


def solve():
    n = II()
    data = LMII()
    dic = {}
    for i in range(n - 1):
        u, v = MII()
        dic[u] = dic.get(u, []) + [v]
        dic[v] = dic.get(v, []) + [u]

    visited = [0] * (n + 1)
    lst = [[1]]
    now_visit = [1]
    visited[1] = 1
    father = {}
    son = {}
    while now_visit:
        next_visit = []
        for x in now_visit:
            for xx in dic[x]:
                if not visited[xx]:
                    father[xx] = x
                    son[x] = son.get(x, []) + [xx]
                    next_visit.append(xx)
                    visited[xx] = 1
        if next_visit:
            lst.append(next_visit)
        now_visit = next_visit
    lst.reverse()

    res = [set() for _ in range(n)]
    for i in range(1, n + 1):
        if i not in son:
            res[i - 1].add(data[i - 1])
    # print(res)
    # print(lst)
    # print(father)
    # print(son)

    def fun(lst, x):
        """
        lst: 一个列表，列表中每个元素都是一个可迭代的集合（或列表），
             表示“这一组”可供选择的数字。
        x: 目标上界（小于等于 x）
        返回: 一个集合 ans，包含所有可能的选数总和（<= x）。
        """
        # 初始只能选空集，总和为 0
        ans = {0}

        # 对每一组可选数字，更新所有可能的总和
        for group in lst:
            new_ans = set(ans)  # 保留“都不选任何数”的那部分
            for s in ans:  # 对已有的每一种“已选总和”
                for num in group:
                    tot = s + num
                    if tot <= x:
                        new_ans.add(tot)
            ans = new_ans
        ans.remove(0)
        return ans

    handle = [0] * (n + 1)
    for x in lst[:-1]:
        for xx in x:
            if not handle[xx]:
                tmp = father[xx]
                need_handle = []
                for y in son[tmp]:
                    handle[y] = 1
                    need_handle.append(res[y - 1])

                # print(need_handle)
                res[tmp - 1] |= fun(need_handle, data[tmp - 1])
                # print(tmp, res[tmp - 1])

    if not res[0]:
        print(0)
    else:
        print(max(res[0]))


T = 1
for _ in range(T):
    solve()

C++

#include"iostream"
#include"math.h"
#include"algorithm"
#include"vector"
#include"string"
#include"unordered_map"
#include"unordered_set"
#include"set"
using namespace std;

unordered_set<int> fun(const vector<unordered_set<int>>& lst, int x) {
	unordered_set<int> ans = { 0 };
	for (const auto& group : lst) {
		unordered_set<int> new_ans;
		for (int s : ans) {
			new_ans.insert(s);  // 记得保留原来的情况（不选当前组）
			for (int num : group) {
				int tot = s + num;
				if (tot <= x) new_ans.insert(tot);
			}
		}
		ans = std::move(new_ans);
	}
	ans.erase(0);
	return ans;
}

int main() {
	int n;
	cin >> n;
	vector<int> data;
	for (int i = 0;i < n;i++) {
		int x;
		cin >> x;
		data.push_back(x);
	}
	unordered_map<int, vector<int>> m;
	for (int i = 0;i < n - 1;i++) {
		int u, v;
		cin >> u >> v;
		m[u].push_back(v);
		m[v].push_back(u);
	}

	vector<int> visited(n + 1, 0);
	vector<vector<int>> lst = { {1} };
	vector<int> now_visit = { 1 };
	visited[1] = 1;
	unordered_map<int, int> father;
	unordered_map<int, vector<int>> son;
	while (!now_visit.empty()) {
		vector<int> next_visit;
		for (auto x : now_visit) {
			for (auto xx : m[x]) {
				if (!visited[xx]) {
					father[xx] = x;
					son[x].push_back(xx);
					next_visit.push_back(xx);
					visited[xx] = 1;
				}
			}
		}
		if (!next_visit.empty()) {
			lst.push_back(next_visit);
		}
		now_visit = next_visit;
	}
	reverse(lst.begin(), lst.end());

	vector<unordered_set<int>> res(n);
	for (int i = 1;i < n + 1;i++) {
		if (son.find(i) == son.end()) {
			res[i - 1].insert(data[i - 1]);
		}
	}

	vector<int> handle(n + 1, 0);
	for (int i = 0; i < (int)lst.size() - 1; i++) {
		for (auto x : lst[i]) {
			if (!handle[x]) {
				int tmp = father[x];
				vector<unordered_set<int>> need_handle;
				for (auto y : son[tmp]) {
					handle[y] = 1;
					need_handle.push_back(res[y - 1]);
				}
				unordered_set<int> merged = fun(need_handle, data[tmp - 1]);
				for (int val : merged) {
					res[tmp - 1].insert(val);
				}
			}
		}
	}

	if (res[0].empty()) {
		cout << 0 << endl;
	}
	else {
		int ans = 0;
		for (auto a : res[0]) {
			if (a > ans)ans = a;
		}
		cout << ans << endl;
	}
}