UVALive 6262 Darts
Description
Consider a game in which darts are thrown at a board. The board is formed by 10 circles with radii 20, 40, 60, 80, 100, 120, 140, 160, 180, and 200 (measured in millimeters), centered at the origin. Each throw is evaluated depending on where the dart hits the board. The score is p points (p
{1, 2,..., 10}) if the smallest circle enclosing or passing through the hit point is the one with radius 20 . (11 - p). No points are awarded for a throw that misses the largest circle. Your task is to compute the total score of a series of n throws.
Input
The first line of the input contains the number of test cases T. The descriptions of the test cases follow:
Each test case starts with a line containing the number of throws n (1
n
106). Each of the next n lines contains two integers x and y (- 200
x, y
200) separated by a space -- the coordinates of the point hit by a throw.
Output
Print the answers to the test cases in the order in which they appear in the input. For each test case print a single line containing one integer -- the sum of the scores of all n throws.
Sample Input
1
5
32 -39
71 89
-60 80
0 0
196 89
Sample Output
29
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
int d(int a,int b)
{
if(*<(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
if(*<(pow(a,)+pow(b,))&&(pow(a,)+pow(b,)))
return ;
return ;
}
int main()
{
int n;
cin>>n;
while(n--)
{
int m,ans=;
cin>>m;
while(m--)
{
int a,b;
scanf("%d%d",&a,&b);
ans+=d(a,b);
}
printf("%d\n",ans);
}
return ;
}
UVALive 6262 Darts的更多相关文章
- UVALive - 4108 SKYLINE[线段树]
UVALive - 4108 SKYLINE Time Limit: 3000MS 64bit IO Format: %lld & %llu Submit Status uDebug ...
- UVALive - 3942 Remember the Word[树状数组]
UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device wit ...
- UVALive - 3942 Remember the Word[Trie DP]
UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here com ...
- 日本DARTS 支撑的一系列应用项目
DARTS是多学科空间科学数据平台,例如天体物理.太阳物理.太阳物理.月球与行星科学和微重力科学.在此数据支撑下,有许多应用. 1.http://wms.selene.darts.isas.jaxa. ...
- 思维 UVALive 3708 Graveyard
题目传送门 /* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离: 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000 ...
- UVALive 6145 Version Controlled IDE(可持久化treap、rope)
题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_ ...
- UVALive 6508 Permutation Graphs
Permutation Graphs Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit ...
- UVALive 6500 Boxes
Boxes Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Pract ...
- UVALive 6948 Jokewithpermutation dfs
题目链接:UVALive 6948 Jokewithpermutation 题意:给一串数字序列,没有空格,拆成从1到N的连续数列. dfs. 可以计算出N的值,也可以直接检验当前数组是否合法. # ...
随机推荐
- mysql只读模式的设置方法与实验【转】
在MySQL数据库中,在进行数据迁移和从库只读状态设置时,都会涉及到只读状态和Master-slave的设置和关系. 经过实际测试,对于MySQL单实例数据库和master库,如果需要设置为只读状态, ...
- 在JAVA中记录日志的十个小建议
JAVA日志管理既是一门科学,又是一门艺术.科学的部分是指了解写日志的工具以及其API,而选择日志的格式,消息的格式,日志记录的内容,哪种消息对应于哪一种日志级别,则完全是基于经验.从过去的实践证明, ...
- ExtJs的Reader
ExtJs的Reader Reader : 主要用于将proxy数据代理读取的数据按照不同的规则进行解析,讲解析好的数据保存到Modle中 结构图 Ext.data.reader.Reader 读取器 ...
- Java 容器的基本概念
java容器类类库的用途时"保存对象",并将其划分为两个不同的概念: 1)Collection(采集).一个独立元素的序列,这些元素都服从一条或多条规则,List必须按照插入的顺序 ...
- Codefroces 628B New Skateboard(数位+思维)
题目链接:http://codeforces.com/contest/628/problem/B 题目大意:给你一段数字串s(1?≤?|s|?≤?3·10^5),求该字符串有多少子串是4的倍数.解题思 ...
- Codeforces 963A Alternating Sum(等比数列求和+逆元+快速幂)
题目链接:http://codeforces.com/problemset/problem/963/A 题目大意:就是给了你n,a,b和一段长度为k的只有'+'和‘-’字符串,保证n+1被k整除,让你 ...
- 打印数据的字节(十六进制)表示-c语言代码
先取数据地址,转换成单字节长度的类型(unsigned char)的指针,然后按照十六进制逐字节打印即可,格式为“%.2x”. sizeof()函数获取数据的字节数. /* $begin show-b ...
- hdu 5122 (2014北京现场赛 K题)
把一个序列按从小到大排序 要执行多少次操作 只需要从右往左统计,并且不断更新最小值,若当前数为最小值,则将最小值更新为当前数,否则sum+1 Sample Input255 4 3 2 155 1 2 ...
- GUC-13 生产者消费者案例
import java.util.concurrent.locks.Condition; import java.util.concurrent.locks.Lock; import java.uti ...
- 用html5实现的flappy-bird
可能网上早就有几个flappy-bird的html5版本啦,到这个时候flappy-bird可能也没有之前那么火了,但是作为一个新手,自己思考,自己动手写一个flappy-bird的demo还是很有成 ...