1.小美的平衡矩阵

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int[][] nums = new int[n][n], sum = new int[n][n];char[] chars;for (int i = 0; i < n; i++) {chars = scanner.next().toCharArray();for (int j = 0; j < n; j++) {nums[i][j] = chars[j] - '0';if ( i != 0 &&j != 0)sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] +nums[i][j];else if (i != 0)sum[i][j] = sum[i - 1][j] + nums[i][j];else if (j != 0)sum[i][j] = sum[i][j - 1] + nums[i][j];else sum[i][j] = nums[i][j];}}int res = 0;int tar = 0;for(int i = 0;i<n;i++){if(i%2==0) System.out.println(0);else{for (int j = i; j < n; j++) {for (int w = i; w < n; w++) {if (j == i && w == i)tar = sum[j][w];else if (j == i)tar = sum[j][w] - sum[j][w - i - 1];else if (w == i)tar = sum[j][w] - sum[j - i - 1][w];else tar = sum[j][w] - sum[j][w - i - 1] - sum[j - i - 1][w] + sum[j - i - 1][w- i - 1];if (tar == (i + 1) * (i + 1) / 2)res++;}}System.out.println(res);res = 0;}}}
}

2. 小美的数组询问

import java.util.Scanner;
import java.io.*;
public class Main {public static void main(String[] args) throws IOException {StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));in.nextToken();int n=(int)in.nval;in.nextToken();int q=(int)in.nval;int[] arr = new int [n];long sum = 0l;long a = 0;for (int i = 0; i < n; i++) {in.nextToken();arr[i] = (int)in.nval;if(arr[i]==0) a++;sum += (long)arr[i];}for (int i = 0; i < q; i++) {in.nextToken();int  left = (int)in.nval;in.nextToken();int right = (int)in.nval;System.out.print(sum+a*left);System.out.print(" ");System.out.println(sum+a*right);}}
}

3.小美的 MT

import java.util.Scanner;
import java.io.*;public class Main {public static void main(String[] args) throws IOException {
//        StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
//        in.nextToken();int n=(int)in.nval;
//        in.nextToken();int q=(int)in.nval;StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));in.nextToken();int n = (int) in.nval;in.nextToken();int k = (int) in.nval;in.nextToken();String string = in.sval;int sum = 0;for(int i = 0;i<string.length();i++){if(string.charAt(i)=='M'||string.charAt(i)=='T'){sum++;}}System.out.println(sum+k>=n?n:sum+k);}
}

import java.util.Scanner;public class Main {static final int maxn = 100010;public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int q = scanner.nextInt();int[] a = new int[maxn];long cnt = 0;long sum = 0;for (int i = 1; i <= n; ++i) {a[i] = scanner.nextInt();if (a[i] == 0) {cnt++;} else {sum += a[i];}}while (q-- > 0) {int l = scanner.nextInt();int r = scanner.nextInt();System.out.println((sum + l * cnt) + " " + (sum + r * cnt));}}
}

4.小美的朋友关系

关键词：并查集、逆序、栈、类、方法重写、集合

这题考到我的智商盲点的，我们需要维护一个并查集来记录朋友关系，这题难点就在于后期会存在遗忘的情况，但是并查集只有合并操作，没有删除操作，由于进行了路径压缩，因此删除的时候难以确定应该修改哪些节点。但是我们可以逆向操作，我们可以逆向遍历查询，遇到删除操作如果是逆序的话则是合并操作，这样就能用并查集进行处理了。确定了大方向后，我们首先读入初始化的边存入数组和集合中，然后存储后期的查询，然后对应后期遗忘的边存入集合方便后续判断。然后才开始初始化关系，注意后期要删除的边不要初始化。然后在存储查询的时候要注意，遗忘中可能包括不是初始化时的操作，是间接关系，是不需要执行并操作的，然后也会出现重复的遗忘，我们要执行加边的是第一次出现的遗忘，因此需要将重复的遗忘从查询中删除。然后要注意重写类的equals方法，传入的参数需要与父类一致，都是Object类，然后hashcode也需要重写，否则集合会判断两者不一样。

import java.util.*;public class Main {static Map<Integer, Integer> fa = new HashMap<>();static Set<Pair> fr = new HashSet<>();static List<Pair> qs = new ArrayList<>();static List<String> ans = new ArrayList<>();static class Pair {int first;int second;int third;Pair(int first, int second) {this.first = first;this.second = second;}Pair(int first, int second, int third) {this.first = first;this.second = second;this.third = third;}@Overridepublic boolean equals(Object o) {if (this == o) return true;if (o == null || getClass() != o.getClass()) return false;Pair pair = (Pair) o;return first == pair.first && second == pair.second && third == pair.third;}@Overridepublic int hashCode() {return Objects.hash(first, second, third);}}static int find(int x) {if (!fa.containsKey(x)) return x;fa.put(x, find(fa.get(x)));return fa.get(x);}static void merge(int x, int y) {x = find(x);y = find(y);if (x != y) {fa.put(x, y);}}public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int m = scanner.nextInt();int q = scanner.nextInt();for (int i = 0; i < m; i++) {int u = scanner.nextInt();int v = scanner.nextInt();fr.add(new Pair(u, v));}for (int i = 0; i < q; i++) {int op = scanner.nextInt();int u = scanner.nextInt();int v = scanner.nextInt();if (op == 1) {fr.remove(new Pair(u, v));}qs.add(new Pair(op, u, v));}Collections.reverse(qs);for (Pair pair : fr) {merge(pair.first, pair.second);}for (Pair pair : qs) {if (pair.first == 1) {merge(pair.second, pair.third);} else {ans.add(find(pair.second) == find(pair.third) ? "Yes" : "No");}}Collections.reverse(ans);for (String s : ans) {System.out.println(s);}}
}

5.小美的区间删除

小美拿到了一个大小为n的数组，她希望删除一个区间后，使得剩余所有元素的乘积末尾至少有k个 0。小美想知道，一共有多少种不同的删除方案？

关键词：数学、前缀和、滑动窗口

这题我只想到了使用前缀和来解决，因此会遇到乘法太大导致溢出的问题，当时还打算使用BigDecimal来解决，原来是自己想的简单了。这题除了前缀和，还考了数学问题，实际上能够得到10的倍数只与2和5的个数相关，其他因子对这个不产生影响。因此我们只需对数组中每个数进行分解，看里面包含多少个2和5，然后用前缀和的方式记录。然后就使用滑动窗口来寻找可以删除的区间。判断条件是这个剩下的区间的2和5的最小值与k进行比较，因为一个2和一个5相乘就是10，那么2和5的最小值就是末尾为零的个数。

import java.util.*;public class MaxCase {public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int k = in.nextInt();long[] pre2 = new long[n+1];long[] pre5 = new long[n+1];int cnt2,cnt5,temp;for(int i=0;i<n;i++){temp = in.nextInt();cnt2=0;cnt5=0;for(int x=temp;x%2==0;x/=2) cnt2++;for(int x=temp;x%5==0;x/=5) cnt5++;pre2[i+1] = pre2[i] + cnt2;pre5[i+1] = pre5[i] + cnt5;}long res = 0;for(int i=0,j=0;i<n;i++){while(j<n){long remain2 = pre2[n] - pre2[j+1] + pre2[i];long remain5 = pre5[n] - pre5[j+1] + pre5[i];if(Math.min(remain2,remain5)<k) break;j++;}res += Math.max(j-i, 0);}System.out.println(res);}
}

垃圾的我写的，我实在不知道哪错了！希望大佬指正！！！备受感谢！！

import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.IOException;public class Main {public static void main(String[] args) throws IOException {StreamTokenizer tokenizer = new StreamTokenizer(new InputStreamReader(System.in));tokenizer.nextToken();        int n = (int) tokenizer.nval;tokenizer.nextToken();        int k = (int) tokenizer.nval;int [] arr = new int[n];long sum = 1l;for (int i = 0; i < arr.length; i++) {tokenizer.nextToken();    arr[i] = (int) tokenizer.nval;sum *=arr[i];}long temp = sum;long result = 0;for(int left = 0; left < arr.length;left++) {for(int right = left;right<arr.length;right++){temp /= arr[right];if(moweizero(temp,k))result++;else {break;}}temp = sum;}System.out.println(result);}public static boolean  moweizero (long sum ,int k){while(k>0){if(sum==0)return false;if(sum%10!=0)return false;sum = sum/10;k--;}return true;}
}