http://acm.zzuli.edu.cn/problem.php?id=1733

转转转~

 

 旋转图像

Time Limit: 1 Sec  Memory Limit: 128 MB

Submit: 297  Solved: 136

Description

Input

Output

Sample Input

2

2 3

111

000

90

3 3

111

101

111

180

Sample Output

01

01

01

111

101

111

 

HINT

#include<stdio.h>

#include<string.h>

#define maxn 55

int ch[maxn][maxn];

void fanzhuan(int n, int m)  //  写一个翻转90度的函数，多次利用

{

    int i, j;

    int hc[maxn][maxn] = {

{0}};

    for(i = 0; i < n; i++)

    {

        for(j = 0; j < m; j++)

            hc[i][j] = ch[i][j];

    }

    memset(ch, 0, sizeof(ch));

    for(i = 0; i < n; i++)

    {

        for(j = 0; j < m; j++)

            ch[j][n-1-i] = hc[i][j]; //  旋转坐标变化

    }

}

int main()

{

    int i, j, t, n, m, angle, a, b, temp;

    scanf("%d", &t);

    while(t--)

    {

        scanf("%d%d", &n, &m);

        for(i = 0; i < n; i++)

            for(j = 0; j < m; j++)

                scanf("%d", &ch[i][j]);

        scanf("%d", &angle);

        a = angle % 180; //  判断行列是否变化

        b = angle % 360;

        if(b == 90)

            fanzhuan(n, m);

        if(b == 180)

        {

            fanzhuan(n, m);

            fanzhuan(m, n);

        }

        if(b == 270)

        {

            fanzhuan(n, m);

            fanzhuan(m, n);

            fanzhuan(n, m);

        }

        if(a == 90)

        {

            temp = m, n = n, n = temp;

        }

        for(i = 0; i < n; i++)

        {

            for(j = 0; j < m; j++)

                printf("%d\n", ch[i][j]);

        }

    }

    return 0;

}