PTA甲 1105. Spiral Matrix (25)
2017-07-25 18:29
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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:
Sample Output:
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;
}
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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