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PTA甲 1105. Spiral Matrix (25)

2017-07-25 18:29 351 查看
This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrixis filled in from the first element at the upper-left corner, then move in a clockwise spiral. The
matrix has m rows and ncolumns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is
the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 104. The
numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.
Sample Input:
12
37 76 20 98 76 42 53 95 60 81 58 93

Sample Output:
98 95 93
42 37 81
53 20 76
58 60 76

这道题题目并不是很难但是在调节的过程中比较麻烦。

ac代码
#include <iostream>

#include <cstdio>

#include <algorithm>

#include <string.h>

#include <cmath>

using namespace std;

bool cmp(int a,int b){

    return a>b;

}

int a[10005];

int matrix[205][205];

int main()

{
int n,r,c,N,M,i,j;
scanf("%d",&n);

for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}

sort(a,a+n,cmp);

int min=100000;

for(c=1;c<=n;c++)
{
for (r=1;r<=c;r++)
{
if(c*r==n)
{
if (c-r<min)
{
min=c-r;
N=r;
M=c;
}
}
}
}

memset(matrix,-1,sizeof(matrix));

int f=0,n0=0,m0=0;

for(i=0;i<n;i++)
{
matrix[m0][n0]=a[i];
if(f==0)
       {
if(n0+1<N&&matrix[m0][n0+1]==-1)
{
n0++;
}
else
{
m0++;
f=1;
}
}
else if(f==1)
{
if(m0+1<M&&matrix[m0+1][n0]==-1)
{
m0++;
}
else
{
n0--;
f=2;
}
}
else if(f==2)
{
if (n0>0&&matrix[m0][n0-1]==-1)
{
n0--;
}
else
{
m0--;
f=3;
}
}
else
{
if(m0>0&&matrix[m0-1][n0]==-1)
{
m0--;
}
else
{
n0++;
f=0;
}
}

}
for(i=0;i<M;i++)
{

for(j=0;j<N-1;j++)
{
printf("%d ",matrix[i][j]);
}
printf("%d",matrix[i][N-1]);

printf("\n");
}

return 0;

}
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