POJ1005 I Think I Need a Houseboat
2017-01-20 18:20
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所拥有的土地以圆的形式扩张,第一个输入是节点个数。
同时又是连续输入,连续输出的情况,再次用到链表。
没有考虑到面积的增大是半圆,并不是精度带来的问题。
关键公式就是1/2∗π∗r2=50∗n
后面在测试代码的时候,发现,自己所采用的的方案,比较会带来误差。
连续输出版本,好几次ac不了在于最后一个句号漏了,可见解题要细心。
不连续输出只需要稍作改动。
同时又是连续输入,连续输出的情况,再次用到链表。
没有考虑到面积的增大是半圆,并不是精度带来的问题。
关键公式就是1/2∗π∗r2=50∗n
后面在测试代码的时候,发现,自己所采用的的方案,比较会带来误差。
double x = current->x,y = current->y; int area_order = 1; double r_square = area_order*100/3.1415926; for(;r_square < x*x+y*y;area_order++) { r_square = area_order*100/3.1415926; cout<<r_square<<endl; }
连续输出版本,好几次ac不了在于最后一个句号漏了,可见解题要细心。
不连续输出只需要稍作改动。
#include <iostream> #include <cmath> using namespace std; struct Node { public: double x; double y; Node* next; }; class List { Node*head; Node* tail; int num; public: //默认构造函数 List(){head = NULL;num = 0;} void pushList(double a,double b);//链表结点的插入 void outputList();//链表结点的输出 }; void List::pushList(double a,double b) { if(head == NULL) { head=(Node*)new Node; head->x = a; head->y = b; head->next = NULL; tail=head; num++; } else { Node* s=(Node*)new Node; s->x = a; s->y = b; s->next = NULL; tail->next = s; tail = s; num++; } } void List::outputList() { Node* current = head; for(int seq = 1;current!=NULL;seq++) { double x = current->x,y = current->y; cout<<"Property "<<seq<<": This property will begin eroding in year "<<ceil(3.1415926*(x*x+y*y)/100.0)<<"."<<endl; current=current->next; } } int main() { int m = 0; List mlist; cin>>m; while(m>0) { m--; double a,b; cin>>a>>b; mlist.pushList(a,b); } mlist.outputList(); cout<<"END OF OUTPUT."<<endl; return 0; }
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