I Think I Need a Houseboat POJ - 1005 解题思路:水题 #include <iostream> #include <cstdio> #include <cmath> using namespace std; int main() { int n; double xx,yy; double rr; double area,r; int year; scanf("%d",&n); ;i<n;i++){ s…

题目大意 在二维坐标内选定一个点,问你当洪水以半圆形扩散且每年扩散50单位,哪一年这个点被被洪水侵蚀? 解法 代码 #include <iostream> #include <cstdio> using namespace std; int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t,cnt=0; double pi=3.14; cin>>t; while(t--) { double…

1.链接地址: http://bailian.openjudge.cn/practice/1005/ http://poj.org/problem?id=1005 2.题目: I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 82376   Accepted: 35628 Description Fred Mapper is considering purchasing…

POJ. 1005 I Think I Need a Houseboat(水 ) 代码总览 #include <cstdio> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <vector> #define nmax using namespace std; const double pi = acos(-1); int m…

    I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 81874   Accepted: 35368 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the la…

I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 85149   Accepted: 36857 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land,…

I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 92090   Accepted: 40012 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land,…

一. 题目 I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 97512   Accepted: 42430 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the…

#include <iostream> using namespace std; const double pi = 3.1415926535; int main() { ;; double x,y; cin >> t; while(t--) { ++time; cin >> x >> y; double area = pi*(x*x + y*y); ; cout<<"Property "<<time<<…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…