第一题
- 字典序最小的 01 字符串
解题思路：
模拟，统计遇到的连续的1的个数记为num，直到遇到0，如果k>=num，直接将第一个1置为0，将遇到的0置为1，否则将第一个1偏置num-k个位置置为0，遇到的0置为1。
原理是遇到的1，基本都要往后移。有多少个k就可以往后移多少个1，而字典序最小又要求我们优先移动前面的
#include <iostream>
#include <string>
using namespace std;int main() {int n, k;cin >> n >> k;string s;cin >> s;int num = 0;int start = 0;for (int i = 0; i < n; i++) {if (s[i] == '0') continue;num = 0;start = i;num++;while (i + 1 < n && s[i+1] != '0') {num++; i++;}if (i + 1 >= n) break;if (k >= num) {s[start] = '0'; s[start + num] = '1';k -= num;i = start;}else {s[start + num - k] = '0'; s[start + num] = '1';k = 0; break;}}cout << s << endl;
}
import java.util.*;class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int k = sc.nextInt();sc.nextLine();String str = sc.nextLine();char[] ch = str.toCharArray();int num = 0;int start = 0;for (int i = 0; i < n; i++) {if (ch[i] == '0') continue;num = 0;start = i;num++;while (i + 1 < n && ch[i+1] != '0') {num++; i++;}if (i + 1 >= n) break;if (k >= num) {ch[start] = '0'; ch[start + num] = '1';k -= num;i = start;}else {ch[start + num - k] = '0'; ch[start + num] = '1';k = 0; break;}}StringBuilder sb = new StringBuilder();for (int i = 0; i < n; i++) sb = sb.append(ch[i]);System.out.println(sb.toString());}
}
第二题
- 数组子序列的排列
解题思路：
先统计从1开始连续的数的个数和每个数在数组中出现的个数，只统计100000以下的数出现的个数，最后计算规律为n1 + n1n2 + n1n2n3 + … + n1n2…*nm
#include<iostream>
#include <vector>
using namespace std;const long long mod = 1000000000 + 7;int main() {int n;cin >> n;vector<long long> nums(n, 0);vector<long long> num(n, 0);for (int i = 0; i < n; i++) {cin >> nums[i];if (nums[i] < 100000)num[nums[i]-1]++;}int b = 0;for (int i = 0; i < n; i++) {if (num[i] == 0) {b = i; break;}}long long sum = 0;for (int i = 0; i < b; i++) {long long sum_in = 1;for (int j = 0; j <= i; j++) {sum_in *= num[j];sum_in %= mod;}sum += sum_in;sum %= mod;}sum %= mod;cout << sum << endl;
}
import java.util.*;class Main {public static void main(String[] args) {long mod = 1000000007L;Scanner sc = new Scanner(System.in);int n = sc.nextInt();sc.nextLine();long[] nums = new long[n];long[] num = new long[n];for (int i = 0; i < n; i++) {nums[i] = sc.nextLong();if (nums[i] < 100000)num[(int)nums[i]-1]++;}int b = 0;for (int i = 0; i < n; i++) {if (num[i] == 0) {b = i; break;}}long sum = 0;for (int i = 0; i < b; i++) {long sum_in = 1;for (int j = 0; j <= i; j++) {sum_in *= num[j];sum_in %= mod;}sum += sum_in;sum %= mod;}sum %= mod;System.out.println(sum);}
}
第三题
- 传送树
解题思路
用dfs扫描一遍邻接表即可
#include <iostream>
#include <list>
#include <vector>
#include <climits>
using namespace std;int dfs(vector<int>& ans, vector<list<int>>& tree, int index) {if (tree[index].size() == 0) {ans[index] = 1;return index;}int ret = INT_MAX;for (int t : tree[index]) ret = min(dfs(ans, tree, t), ret);ans[index] = ans[ret] + 1;return min(index, ret);
}int main() {int n; cin >> n;vector<list<int>> tree(n, list<int>(0));vector<int> ans(n, 0);int u, v;for (int i = 0; i < n - 1; i++) {cin >> u >> v;tree[u-1].push_back(v-1);}dfs(ans, tree, 0);for (int index = 0; index < n; index++) {cout << ans[index] << ' ';}cout << endl;
}
import java.util.*;class Main {public static int dfs(int[] ans, List<Integer>[] tree, int index) {if (tree[index].size() == 0) {ans[index] = 1;return index;}int ret = Integer.MAX_VALUE;for (int t : tree[index]) ret = Math.min(dfs(ans, tree, t), ret);ans[index] = ans[ret] + 1;return Math.min(index, ret);}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();List<Integer>[] tree = new List[n];for (int i = 0; i < n; i++) tree[i] = new ArrayList<>();int[] ans = new int[n];int u, v;for (int i = 0; i < n - 1; i++) {u = sc.nextInt();v = sc.nextInt();tree[u-1].add(v-1);}dfs(ans, tree, 0);for (int index = 0; index < n; index++) {System.out.print(ans[index] + " ");}System.out.println();}
}
