Who Gets the Most Candies?

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 262144/131072K (Java/Other)
Total Submission(s) : 15   Accepted Submission(s) : 7
Problem Description

N children are sitting in a circle to play a game.

The children are numbered from 1 to N in clockwise order. Each of them has a card with a non-zero integer on it in his/her hand. The game starts from the K-th child, who tells all the others the integer on his card and jumps out of the circle. The integer on his card tells the next child to jump out. Let A denote the integer. If A is positive, the next child will be the A-th child to the left. If A is negative, the next child will be the (−A)-th child to the right.

The game lasts until all children have jumped out of the circle. During the game, the p-th child jumping out will get F(p) candies where F(p) is the number of positive integers that perfectly divide p. Who gets the most candies?

 
Input
There are several test cases in the input. Each test case starts with two integers N (0 < N ≤ 500,000) and K (1 ≤ KN) on the first line. The next N lines contains the names of the children (consisting of at most 10 letters) and the integers (non-zero with magnitudes within 108) on their cards in increasing order of the children’s numbers, a name and an integer separated by a single space in a line with no leading or trailing spaces.
 
Output

Output one line for each test case containing the name of the luckiest child and the number of candies he/she gets. If ties occur, always choose the child who jumps out of the circle first.

 
Sample Input
4 2
Tom 2
Jack 4
Mary -1
Sam 1
 
Sample Output
Sam 3
/*题意:N个人围成一圈第一个人跳出圈后会告诉你下一个谁跳出来跳出来的人
(如果他手上拿的数为正数,从他左边数A个,反之,从他右边数A个)
跳出来的人所得到的糖果数量和他跳出的顺序有关 所得的糖果数为 (假设他是第k个跳出的) 则他得到的糖数为k能被多少个数正数
比如说 k = 6 ; 6 = 1*2*3*6 所以他得到的糖数为4; 思路:线段数 先算出N个人中,是第几个人(id)跳出来得到的糖果最多。然后模拟id遍 长到第id个人的name
线段树的结点中保存该区间内还剩多少人,每次update 删除一个人。*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#define maxn 500004
int n,id;
struct node
{
int left,right;
int num;
};
node tree[*maxn];
struct data
{
int val;
char name[];
} boy[maxn]; int ans[maxn];
void build(int left,int right,int i)
{
tree[i].left =left;
tree[i].right =right;
tree[i].num =right-left+;
if(tree[i].left ==tree[i].right )
return ;
build(left,(left+right)/,*i);
build((left+right)/+,right,*i+);
}
int insert(int i,int x)
{
tree[i].num--;
if(tree[i].left==tree[i].right)
return tree[i].left;
if(x<=tree[i].num)
return insert(*i,x);
else
return insert(*i+,x-tree[*i].num);
}
void count_ans()//这个函数最重要。
{
int i,max,j;
memset (ans,,sizeof(ans)); //计算ans
for (i = ; i <= n; i ++)
{
ans[i] ++;
for (j=*i;j<= n;j+= i)
ans[j] ++;
}
max = ans[];
id = ;
for (i = ; i <= n; i ++) //找出第几个人跳出获得的糖最多
if (ans[i] > max)
{
max = ans[i];
id = i;
}
}
int main ()
{
int i,k,mod;
while (~scanf ("%d %d",&n,&k))
{
count_ans();
for (i = ; i <= n; i ++)
scanf ("%s %d",boy[i].name,&boy[i].val);
build(,n,);
mod =tree[].num;
int pos = ;
boy[].val = ;
n = id;
while (n --)
{
if (boy[pos].val > ) //k表剩余的人中从左起第k中出队(PS:k的求法是看别人的)
k = ((k + boy[pos].val - )%mod + mod)%mod + ;
else
k = ((k + boy[pos].val - )%mod + mod)%mod + ;
pos = insert(,k);
mod = tree[].num;
}
printf ("%s %d\n",boy[pos].name,ans[id]);
}
return ;
}

HDU- Who Gets the Most Candies?的更多相关文章

  1. 【HDU 6126】Give out candies 最小割

    题意 有$n​$个小朋友,给每个人分$1~m​$个糖果,有k个限制 限制形如$(x,y,z)​$ 表示第$x​$个人分到的糖数减去第$y​$个人分到的糖数不大于$z​$,给第$i​$个人$j​$颗糖获 ...

  2. HDU 6126.Give out candies 最小割

    Give out candies Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Other ...

  3. hdu 6126 Give out candies

    hdu 6126 Give out candies(最小割) 题意: 有\(n\)个小朋友,标号为\(1\)到\(n\),你要给每个小朋友至少\(1\)个且至多\(m\)个的糖果.小朋友们共提出\(k ...

  4. Hdu 5407 CRB and Candies (找规律)

    题目链接: Hdu 5407 CRB and Candies 题目描述: 给出一个数n,求lcm(C(n,0),C[n,1],C[n-2]......C[n][n-2],C[n][n-1],C[n][ ...

  5. HDU 6085 - Rikka with Candies | 2017 Multi-University Training Contest 5

    看了标程的压位,才知道压位也能很容易写- - /* HDU 6085 - Rikka with Candies [ 压位 ] | 2017 Multi-University Training Cont ...

  6. HDU 5127 Dogs' Candies

    Dogs' Candies Time Limit: 30000/30000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) T ...

  7. 2017ACM暑期多校联合训练 - Team 5 1001 HDU 6085 Rikka with Candies (模拟)

    题目链接 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, s ...

  8. HDU 6085 Rikka with Candies(bitset)

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=6085 [题目大意] 给出一个数组a一个数组b,以及询问数组c, 问对于每个c有多少对a%b=c,答 ...

  9. 2017多校第5场 HDU 6085 Rikka with Candies bitset

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6085 题意:存在两个长度为n,m的数组A,B.有q个询问,每个询问有一个数字k,可以得到Ai%Bj=k ...

  10. HDU 5407(2015多校10)-CRB and Candies(组合数最小公倍数+乘法逆元)

    题目地址:pid=5407">HDU 5407 题意:CRB有n颗不同的糖果,如今他要吃掉k颗(0<=k<=n),问k取0~n的方案数的最小公倍数是多少. 思路:首先做这道 ...

随机推荐

  1. mysql 账户操作

    1.授权 mysql> grant 权限1,权限2,…权限n on 数据库名称.表名称 to 用户名@用户地址 identified by ‘连接口令’; 权限1,权限2,…权限n代表selec ...

  2. (转) UIALertView的基本用法与UIAlertViewDelegate对对话框的事件处理方法

    首先,视图控制器必须得实现协议UIAlertViewDelegate中的方法,并指定delegate为self,才能使弹出的Alert窗口响应点击事件. 具体代码如下: #import <UIK ...

  3. CentOS 6.4 64位 安装 jdk 6u45

    准备: 1.下载历史版本jdk 地址: http://java.sun.com/products/archive/ 下载的版本 jdk-6u45-linux-x64-rpm.bin  Linux x6 ...

  4. 国内各大互联网公司UED(用户体验设计)团队博客介绍

     UED是什么UED = user experience design,用户体验设计.UED的通常理解,就是“我们做的一切都是为了呈现在您眼前的页面”.UED团队包括:交互设计师(Interactio ...

  5. ImageList半透明,Alpha通道bug处理。

    由于ImageList的先天障碍,对alpha通道支持不好.虽然到xp有所改善,但瑕疵依然存在. 通过reflactor发现ImageList通过windows api来进行读写的.写入数据时会对原始 ...

  6. qt 5 小练习 简易画板

    如何在窗口上画线?用一根根线来拼凑图案呢? 想必大家都知道点的集合是线,而线的集合就是很多线啦,用线的集合我们能拼凑出许许多多的图案.于是我就要记录自己跟着老师的学习之路啦: 既然有集合的话,势必要用 ...

  7. C#网页自动登录和提交POST信息的多种方法(转)

    网页自动登录和提交POST信息的核心就是分析网页的源代码(HTML),在C#中,可以用来提取网页HTML的组件比较多,常用的用WebBrowser.WebClient.HttpWebRequest这三 ...

  8. JavaScript自学代码--(三)

    //通过 id 查找 HTML 元素 var x = document.getElementById("demo"); //通过标签名查找 HTML 元素 //本例查找 id=&q ...

  9. Painting The Wall 期望DP Codeforces 398_B

    B. Painting The Wall time limit per test 1 second memory limit per test 256 megabytes input standard ...

  10. matlab中 hold on 与hold off的用法

    matlab中 hold on 与hold off的用法 hold on 是当前轴及图形保持而不被刷新,准备接受此后将绘制 hold off 使当前轴及图形不在具备被刷新的性质 hold on 和ho ...