2321.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#define debug 0
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define INF 30000
#define NMAX 11
double m[NMAX][NMAX];
int a[NMAX*NMAX];
int b[NMAX][NMAX];
double p[NMAX][NMAX];
double ans[NMAX]={0},t[NMAX]={0};
int main()
{

#if debug 	
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	int i=0,j;
	while(scanf("%d",&a[i++])!=EOF);
	int n=(int)sqrt(i);
	int k=0;
	
	for(i=0;i<n;i++)
	{
		double sum=0;
		for(j=0;j<n;j++)
		{
			sum+=a[k];
			b[i][j]=a[k];
			k++;
		}
		for(j=0;j<n;j++)
			p[i][j]=b[i][j]/sum;
	}
	t[0]=1;
	ans[0]=1;
	for(i=0;i<n;i++)
	{
		printf("%.5f",ans[i]);
		printf(" ");
	}
	printf("\n");

	for(k=2;k<=10;k++)
	{
		for(j=0;j<n;j++)
		{
			ans[j]=0;
			for(i=0;i<n;i++)
			{
				ans[j]+=t[i]*p[i][j];
				
			}
			printf("%.5f ",ans[j]);
		}			
		memcpy(t,ans,sizeof(t));
		printf("\n");
	}
		

	
#if debug
	fclose(stdin);
	fclose(stdout);
#endif
	return 1;
}