1105. Spiral Matrix (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, 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:

1237 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 9342 37 8153 20 7658 60 76

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;#define mx 10010int ans[mx];int dp[mx][mx];bool cmp(int a,int b){return a>b;}int main(){int N;int n,m;scanf("%d",&N);memset(ans,0,sizeof(ans));for(int i=0;i<N;i++){scanf("%d",&ans[i]);}if(N==1){ //不能忽略，否则某个case通过不了printf("%d",ans[0]);return 0;}sort(ans,ans+N,cmp);//int idx,minnum=99999,num,x,y;/*for(int j=2;j<=N;j++) //按这个方法求比较麻烦，但是也可行{if(N%j!=0)continue;idx=N/j;if(idx>j){x=idx;y=j;}else{x=j;y=idx;}num=x-y;if(num<minnum){minnum=num;m=x;n=y;}}*///printf("%d %d",n,m);m=(int)ceil(sqrt(1.0*N));while(N%m!=0){m++;}n=N/m;int D=n,B=m,L=1,U=1,now=0;int i=1,j=1;while(now<N){while(now<N&&j<D){dp[i][j]=ans[now++];j++;}while(now<N&&i<B){dp[i][j]=ans[now++];i++;}while(now<N&&j>L){dp[i][j]=ans[now++];j--;}while(now<N&&i>U){dp[i][j]=ans[now++];i--;}U++,L++,D--,B--;i++,j++;if(now==N-1){//最后一个数据单独处理，否则某个case通过不了dp[i][j]=ans[now++];}}for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){if(j!=1)printf(" ");printf("%d",dp[i][j]);}printf("\n");}//system("pause");return 0;}