牛客刷题-Day18

牛客刷题-Day18

今日刷题：\(1001-1010\)

1001 模拟 · 例1-字符串展开

解题思路

模拟，注意细节就可以。

C++ 代码

#include <bits/stdc++.h>
#include <sstream>
using namespace std;
const int N = 20, M = 110;
typedef long long LL;

int p1, p2, p3;
string str;

int toLower(char c) {
	if (c >= 'A' && c <= 'Z')
		return c - 'A' + 'a';
	return c;
}

int toUpper(char c) {
	if (c >= 'a' && c <= 'z')
		return c - 'a' + 'A';
	return c;
}

bool isDigit(char c) {
	return c >= '0' && c <= '9';
}

bool isUpper(char c) {
	return c >= 'A' && c <= 'Z';
}

bool isLower(char c) {
	return c >= 'a' && c <= 'z';
}

int main() {
	cin >> p1 >> p2 >> p3 >> str;
	int i = 0;
	while (i < str.size()) {
		if (str[i] != '-') { // 字母直接输出
			cout << str[i++];
		} else {
			if (str[i + 1] - str[i - 1] == 1) { // 后继忽略 -
				i++;
				continue;
			}
			if (!(isUpper(str[i - 1]) && isUpper(str[i + 1]) 
				|| isLower(str[i - 1]) && isLower(str[i + 1]) 
				|| isDigit(str[i - 1]) && isDigit(str[i + 1]))) { // 两侧不同
				cout << str[i++];
				continue;
			}
			if (str[i - 1] - str[i + 1] >= 0) { // 非递增直接输出
				cout << str[i++];
				continue;
			}
			if (str[i + 1] - str[i - 1] > 0) { // 递增
				if (isDigit(str[i - 1])) { // 数字
					if (p1 != 3) { // 填充数字
						if (p3 == 1) {
							for (char c = str[i - 1] + 1; c < str[i + 1]; c++)
								for (int i = 0; i < p2; i++)
									cout << c;
						} else {
							for (char c = str[i + 1] - 1; c > str[i - 1]; c--)
								for (int i = 0; i < p2; i++)
									cout << c;
						}
					} else { // 填充 *
						for (char c = str[i - 1] + 1; c < str[i + 1]; c++)
							for (int i = 0; i < p2; i++)
								cout << '*';
					}
					i++;
					continue;
				}
				if (p1 == 1) { // 填充小写字母
					if (p3 == 1) {
						for (char c = toLower(str[i - 1] + 1); c < toLower(str[i + 1]); c++)
							for (int i = 0; i < p2; i++)
								cout << c;
					} else {
						for (char c = toLower(str[i + 1] - 1); c > toLower(str[i - 1]); c--)
							for (int i = 0; i < p2; i++)
								cout << c;
					}
				} else if (p1 == 2) { // 填充大写字母
					if (p3 == 1) {
						for (char c = toUpper(str[i - 1] + 1); c < toUpper(str[i + 1]); c++)
							for (int i = 0; i < p2; i++)
								cout << c;
					} else {
						for (char c = toUpper(str[i + 1] - 1); c > toUpper(str[i - 1]); c--)
							for (int i = 0; i < p2; i++)
								cout << c;
					}
				} else { // 填充 *
					for (char c = str[i - 1] + 1; c < str[i + 1]; c++)
							for (int i = 0; i < p2; i++)
								cout << '*';
				}
				i++;
			}
		}
	}
	return 0;
} 

1002 模拟 · 例2-多项式输出

解题思路

模拟，要考虑各种特殊情况。

C++ 代码

#include <bits/stdc++.h>
#include <sstream>
using namespace std;
typedef long long LL;
const int N = 110, M = 30;

int n;
int a[N];
string item[N];

int main() {
	cin >> n;
	for (int i = 0; i <= n; i++)
		cin >> a[i];
	for (int i = 0; i <= n; i++) {
		if (a[i] == 0) {
			item[i] = "";
		} else {
			if (n - i == 0) {
				item[i] = "" + to_string(a[i]);
			} else if (n - i == 1) {
				if (abs(a[i]) != 1) {
					item[i] = "" + to_string(a[i]) + "x";
				} else {
					if (a[i] == -1) {
						item[i] = "-x";
					} else {
						item[i] = "x";
					}
				}
			} else {
				if (abs(a[i]) != 1) {
					item[i] = "" + to_string(a[i]) + "x^" + to_string(n - i);
				} else {
					if (a[i] == -1) {
						item[i] = "-x^" + to_string(n - i);
					} else {
						item[i] = "x^" + to_string(n - i);
					}
				}
			}
		}
	}
	string res = "";
	for (int i = 0; i <= n; i++) {
		if (res.size() == 0) {
			res = res + item[i];
		} else {
			if (item[i].size() > 0 && item[i][0] != '-') {
				res = res + "+" + item[i];
			} else {
				res = res + item[i];
			}
		}
	}
	if (res.size() == 0)
		res = "0";
	cout << res << endl;
	return 0;
} 

1005 枚举 · 例2-最大正方形

解题思路

枚举，枚举相邻的两个点，然后根据边长度相等和平行垂直的性质，计算余下两个点。

C++ 代码

#include <bits/stdc++.h>
using namespace std;
const int N = 110;

int n;
char g[N][N];
int x[4], y[4];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%s", g[i] + 1);
    int len = 0;
    for (int x1 = 1; x1 <= n; x1++)
        for (int y1 = 1; y1 <= n; y1++)
            if (g[x1][y1] == '#')
                for (int x2 = 1; x2 <= n; x2++)
                    for (int y2 = 1; y2 <= n; y2++)
                        if (g[x2][y2] == '#') {
                            int x3 = (x1 + x2 + y2 - y1) / 2, y3 = (y1 + y2 + x1 - x2) / 2;
                            int x4 = (x1 + x2 + y1 - y2) / 2, y4 = (y1 + y2 + x2 - x1) / 2;
                            if (x3 >= 1 && x3 <= n && y3 >= 1 && y3 <= n 
                                && x4 >= 1 && x4 <= n && y4 >= 1 && y4 <= n) {
                                if (g[x3][y3] == '#' && g[x4][y4] == '#') {
                                    int t = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
                                    if (t > len) {
                                        len = t;
                                        x[0] = x1, x[1] = x2, x[2] = x3, x[3] = x4;
                                        y[0] = y1, y[1] = y2, y[2] = y3, y[3] = y4;
                                    }
                                }
                            }
                        }
    for (int i = 0; i < 4; i++)
        printf("%d %d\n", x[i], y[i]);
    return 0;
}

1010 枚举 · 例8扩展-校门外的树：hard

解题思路

区间合并。

C++ 代码

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
typedef pair<int, int> PII;

int n, len;
vector<PII> a, c; 

void merge(vector<PII> a, vector<PII> &c) {
    int st = -1, ed = -1;
    for (int i = 0; i < a.size(); i++) {
        if (ed < a[i].first) {
            if (st != -1)
                c.push_back(make_pair(st, ed));
            st = a[i].first, ed = a[i].second;
        } else {
            ed = max(ed, a[i].second);
        }
    }
    if (st != -1)
        c.push_back(make_pair(st, ed));
}

int main() {
    scanf("%d%d", &len, &n);
    for (int i = 1; i <= n; i++) {
        int l, r;
        scanf("%d%d", &l, &r);
        a.push_back(make_pair(l, r));
    }
    sort(a.begin(), a.end());
    merge(a, c);
    int cnt = 0;
    for (int i = 0; i < c.size(); i++) {
        // printf("%d %d\n", c[i].first, c[i].second);
        cnt += (c[i].second - c[i].first + 1);
    }
    printf("%d\n", len + 1 - cnt);
    return 0;
}