第一题：小美的平衡矩阵
注意in.nextLine()和in.next()
import java.util.Scanner;public class Main {static final int maxn = 210;public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();char[][] a = new char[maxn][maxn];// int[][] b = new int[maxn][maxn];int[][] dp = new int[maxn][maxn];in.nextLine(); // 不加就报错！！！// 这是因为前面读取的是整数，后面是nextline取字符串，会先取后面的换行符。// for (int i = 0; i < n; ++i) {for (int i = 1; i <= n; ++i) {String line = in.nextLine();//String line = in.next(); // 如果这么写就不需要in.nextLine();了// 因为如果读取到的标记之间有空格或换行符，则 .next() 方法会将其忽略。返回的是标记（token）字符串。for (int j = 1; j <= n; ++j) {// a[i][j] = line.charAt(j);a[i][j] = line.charAt(j - 1);}}// for (int i = 0; i < n; ++i) {// for (int j = 0; j < n; ++j) {for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n; ++j) {// b[i][j] = a[i][j] == '1' ? 1 : 0;int value = a[i][j] == '1' ? 1 : 0;dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + value;}}for (int len = 1; len <= n; ++len) {if (len % 2 == 1) System.out.println(0);else {int sum = 0;for (int i = len; i <= n; ++i) {for (int j = len; j <= n; ++j) {int num = dp[i][j] - dp[i - len][j] - dp[i][j - len] + dp[i - len][j - len];// if (num == len)if (num * 2 == len * len) sum++;}}System.out.println(sum);}}}
}
第二题：小美的数组询问
注意注释的所有内容！！！都可能让代码超时！！！
- 数组大小预设
- 不用多余读取空行
- System.out.println 通常会比 System.out.printf 更快，后者常用于同时输出不同类型的数据
import java.util.Scanner;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {static final int maxn = 100010;public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int q = in.nextInt();// in.nextLine();// int[] a = new int[n];int[] a = new int[maxn];long sum = 0, num_0 = 0;for (int i = 0; i < n; ++i) {a[i] = in.nextInt();sum += a[i];if (a[i] == 0) num_0++;}// in.nextLine();while (q-- > 0) {int l = in.nextInt();int r = in.nextInt();// System.out.printf("%d %d\n", sum + l * num_0, sum + r * num_0);System.out.println((sum + l * num_0) + " " + (sum + r * num_0));}}
}
第三题：小美的MT
注意ACM模式答案要打印输出，不要return
import java.util.Scanner;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int k = scanner.nextInt();scanner.nextLine();String str = scanner.nextLine();int ans = 0, len = str.length();for (int i = 0; i < len; ++i) {if (str.charAt(i) == 'M' || str.charAt(i) == 'T') ans++;}// return ans + k > len ? len : ans + k;ans = Math.min(ans + k, len);System.out.println(ans);}
}
第四题：小美的朋友关系 3/30
并查集
// import java.util.Scanner;
import java.util.*;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {static final int N = 100010;static List<String> res = new ArrayList<>();static int[] f = new int[N];public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int m = in.nextInt();int q = in.nextInt();// boolean[][] relations = new boolean[n][n];for (int i = 0; i < N; ++i) {f[i] = i;}Set<int[]> edge = new HashSet<>();while (m-- > 0) {int a = in.nextInt();int b = in.nextInt();edge.add(new int[] {a, b});}Set<int[]> del_edge = new HashSet<>();List<int[]> ops = new ArrayList<>();int temp = q;while (temp-- > 0) {int op = in.nextInt();int a = in.nextInt();int b = in.nextInt();if (op == 1) {del_edge.add(new int[] {a, b});del_edge.add(new int[] {b, a});}ops.add(new int[] {op, a, b});// test// for (int i = 0; i < n; ++i) {// for (int j = 0; j < n; ++j) {// System.out.print(relations[i][j] + " ");// }// System.out.println();// }}// 建立并查集for (int[] e : edge) {boolean flag = true;for (int[] de : del_edge) {if (e[0] == de[0] && e[1] == de[1]) {flag = false;break;}}if (flag) merge(e[0], e[1]);}for (int i = q - 1; i >= 0; --i) {int op = ops.get(i)[0], a = ops.get(i)[1], b = ops.get(i)[2];if (op == 1) {merge(a, b);} else {if (find(a) == find(b)) res.add("Yes");else res.add("No");}}for (int i = res.size() - 1; i >= 0; --i) {System.out.println(res.get(i));}}public static int find(int x) {if (x != f[x]) {f[x] = find(f[x]);}return f[x];}public static void merge(int x, int y) {int fx = find(f[x]), fy = find(f[y]);f[fx] = fy;}
}
第五题：小美的区间删除
import java.util.Scanner;// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {static final int maxn = 100010;static int[] a = new int[maxn], pre2 = new int[maxn], pre5 = new int[maxn];// pre[i]表示以i结尾之前所有字符含2或5的个数，不包括第i个static int n, k;static int total2 = 0, total5 = 0;public static void main(String[] args) {Scanner in = new Scanner(System.in);n = in.nextInt();k = in.nextInt();// System.out.println(n+" "+k);for (int i = 0; i < n; ++i) {a[i] = in.nextInt();int cnt2 = cal(a[i], 2), cnt5 = cal(a[i], 5);total2 += cnt2;total5 += cnt5;pre2[i + 1] = pre2[i] + cnt2;pre5[i + 1] = pre5[i] + cnt5;// System.out.println(cnt2+" "+cnt5);// System.out.println(pre2[i + 1]+" "+pre2[i]);}int res = 0;for (int i = 0, j = 0; i < n; ++i) {while (j < n) {// int cnt2 = pre2[j + 1] - pre2[i + 1];// int cnt5 = pre5[j + 1] - pre5[i + 1];int cnt2 = pre2[j + 1] - pre2[i];int cnt5 = pre5[j + 1] - pre5[i];int remain2 = total2 - cnt2, remain5 = total5 - cnt5;if (remain2 >= k && remain5 >= k) j++;else break;}// res += Math.max(j - i + 1, 0);res += Math.max(j - i, 0);}System.out.println(res);}public static int cal(int x, int mod) {int cnt = 0;while (x != 0) {if (x % mod == 0) cnt++;else break;x /= mod;}return cnt;}
}
——Simulink布线技巧)
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