codeforces 405A Gravity Flip (模拟题，简单)

1、http://codeforces.com/problemset/problem/405/A

2、题目大意：

给出一个矩阵，里边方格里有东西，然后将这个矩阵向右到，那么重力作用下就都向右走，求最后每一列有多少个

题目不是很难，模拟也较简单

3、题目：

A. Gravity Flip

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.

There are n columns of toy cubes in the box arranged in a line. Thei-th column contains a_i cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.

Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of then columns after the gravity switch!

Input

The first line of input contains an integer n (1 ≤ n ≤ 100), the number of the columns in the box. The next line containsn space-separated integer numbers. The i-th number a_i (1 ≤ a_i ≤ 100) denotes the number of cubes in thei-th column.

Output

Output n integer numbers separated by spaces, where thei-th number is the amount of cubes in the i-th column after the gravity switch.

Sample test(s)

Input

43 2 1 2

Output

1 2 2 3 

Input

32 3 8

Output

2 3 8 

Note

The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.

In the second example case the gravity switch does not change the heights of the columns.

4、AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 105int a[N][N];int b[N];int main(){    int n,m;    while(scanf("%d",&n)!=EOF)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        int maxx=-1;        for(int i=1; i<=n; i++)        {            scanf("%d",&b[i]);            if(maxx<b[i])                maxx=b[i];            for(int j=1; j<=b[i]; j++)                a[j][i]=1;        }        for(int i=1; i<=maxx; i++)        {            for(int j=n; j>=1; j--)            {                if(a[i][j]==1)                {                    for(int k=j+1; k<=n; k++)                    {                        if(a[i][k]==0 && (a[i][k+1]==1 || k==n))                        {                            a[i][j]=0;                            a[i][k]=1;                            b[k]++;                            b[j]--;                            break;                        }                        else if(a[i][k]==1)                        break;                    }                }            }        }        for(int i=1;i<=n;i++)        printf("%d ",b[i]);        printf("\n");    }    return 0;}/*43 2 1 232 3 8*/