> #include <iostream>
> #include <cmath>
> using namespace std;
> 
> #define PI 3.141593
> 
> int main()
> {
> 	int n;//how many sets
> 	float* x;
> 	float* y;
> 	float* radius;
> 	int i;//year
> 	int j=0;
> 	int count=0;
> 	int *result;//存放结果，为了最终按顺序打印
> 	float r;//每年淹没的半径
> 	cin>>n;
> 	x = (float*)malloc(sizeof(float)*n);
> 	y = (float*)malloc(sizeof(float)*n);//数目不确定，只能用指针
> 	radius = (float*)malloc(sizeof(float)*n);
> 
> 	result = (int*)malloc(sizeof(int)*n);
> 
> 	for(i=0;i<n;i++)//有多少n就要有多少对xy
> 	{
> 		cin>>*(x+i)>>*(y+i);
> 		*(radius+i) = sqrt((*(x+i))*(*(x+i)) + (*(y+i))*(*(y+i)));
> 		*(result+i)=0;
> 	}
> 
> 	for(i=1;;i++)//按年份往上算
> 	{
> 		r = sqrt(50.0*2*i/PI);
> 		for(j=0;j<n;j++)
> 		{
> 			if(*(radius+j)<=r && *(result+j)==0)//第j组被淹				
> 			{				
> 				*(result+j) = i;
> 				count++;
> 			}
> 		}				
> 		if(count==n)
> 			break;
> 	}
> 
> 	for (i=0;i<n;i++)
> 		cout<<"Property "<<i<<": This property will begin eroding in year "<<*(result+i)<<"."<<endl;
> 	cout<<"END OF OUTPUT.";
> 
> 	return 0;
> }

• 想复杂了，纯粹水题 (399) liujunjun 2012-04-20 22:33:46