Educational Codeforces Round 20总结

A. Maximal Binary Matrix

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.

One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.

If there exists no such matrix then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).

Output

If the answer exists then output resulting matrix. Otherwise output -1.

Examples

input

2 1

output

1 0 0 0 

input

3 2

output

1 0 0 0 1 0 0 0 0 

input

2 5

output

-1

题意：
给你一个矩阵 和一个数字k问您是否能完全填充 
填充规则按照字典序（从上到下）和中轴线对称的方法

解：从上到下遍历处理一下就over

import java.util.Scanner;public class Main {private static final int Max = (int) (1e2 + 10);private static int n, k;private static int[][] vis;public static void main(String[] args) {InitData();GetAns();}private static void InitData() {Scanner cin = new Scanner(System.in);n = cin.nextInt();k = cin.nextInt();vis = new int[Max][Max];}private static void GetAns() {if (n * n < k) {System.out.println(-1);} else {for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (vis[i][j] == 1) {continue;}if (i == j && k != 0) {vis[i][j] = 1;k--;} else if (k >= 2) {vis[i][j] = 1;vis[j][i] = 1;k -= 2;}}}if (k != 0) {System.out.println(-1);} else {for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (j == 0) {System.out.print(vis[i][j]);} else {System.out.print(" " + vis[i][j]);}}System.out.println();}System.out.println();}}}}

B. Distances to Zero

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given the array of integer numbers a0, a1, ..., an - 1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — length of the array a. The second line contains integer elements of the array separated by single spaces ( - 109 ≤ ai ≤ 109).

Output

Print the sequence d0, d1, ..., dn - 1, where di is the difference of indices between i and nearest j such that aj = 0. It is possible that i = j.

Examples

input

92 1 0 3 0 0 3 2 4

output

2 1 0 1 0 0 1 2 3 

input

50 1 2 3 4

output

0 1 2 3 4 

input

75 6 0 1 -2 3 4

output

2 1 0 1 2 3 4 

题意：n个元素，求每个数到离它最近的0的距离

解：我这里采用的是二分答案，就是求每个点到0最短的距离

import java.util.Scanner;public class Main {private static final int Max = (int) (2e5 + 10);private static int[] A;private static int d;private static int[] ans;private static int cnt;private static int n;public static void main(String[] args) {InitData();//for (int i = 0; i < cnt; i++) {//System.out.print(ans[i] + "  ");//}for (int i = 1; i <= n; i++) {d = EF(i);if (i == 1) {System.out.print(d);} else {System.out.print(" " + d);}}System.out.println();}private static void InitData() {Scanner cin = new Scanner(System.in);A = new int[Max];ans = new int[Max];cnt = 0;n = cin.nextInt();for (int i = 1; i <= n; i++) {A[i] = cin.nextInt();if (A[i] == 0) {ans[cnt++] = i;}}}// 二分查找，查找一个最接近K的值private static int EF(int k) {int start = 0;int end = cnt - 1;if (k > ans[end]) {return Math.abs(k - ans[end]);} else if (k < ans[start]) {return Math.abs(k - ans[start]);}while (start <= end) {int midle = (end - start) / 2 + start;if (ans[midle] == k) {return Math.abs(ans[midle] - k);} else if (ans[midle] > k) {end = midle - 1;} else {start = midle + 1;}}return Math.min(Math.abs(k - ans[start]), Math.abs(ans[end] - k));}}