POJ 2886 Who Gets the Most Candies?

Who Gets the Most Candies?

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 10234 Accepted: 3177Case Time Limit: 2000MS

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 ≤ K ≤ N) 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 2Tom 2Jack 4Mary -1Sam 1

Sample Output

Sam 3

Source

POJ Monthly--2006.07.30, Sempr

解题思路：首先确定反素数的概念，反素数是指约数个数大于所有小于它的数的约数的个数的数字，那么对于1-n个孩子，我们只要找出其中最大的反素数记为flag，其约数的个数就是答案，打出反素数表和反素数的约数表，然后就是用线段树模拟约瑟夫环，线段树中保存区间内剩下的人数，明确两个值，一是pos，这个是线段树中点的位置，也是一开始各个人的位置，另一个是k，这个是人的相对位置，比如原来是排第三的人现在前面两个人都被T了那么他现在的相对位置，就是k值就是1，考虑k值如何计算，如果一个人的号码牌是正数，那么他被T出后下一个人的相对位置就是k=k+num-1，如果一个人的号码牌是负数，那么他被T出后下一个人的相对位置k=k-num，但是考虑到在环上需要取模，最终k值计算就是正数k=((k-1+num-1)+mod)%mod+1；负数k=((k-1-num)+mod)%mod+1；利用k值找出人在线段树中绝对位置，一直更新flag次就是答案;PS:题目中说出现循环的情况就输出第一个出列的孩子，我也写上了，但是似乎这种情况是不可能出现的？PS2：反素数的说法似乎存疑，维基百科上的反素数似乎另有所指，此处的反素数的正规说法应该是高合成数，但是大家也都是这么叫的，所以就这样吧。。。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#define Max 500005#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;int tree[Max<<2];int RPrime[]={//反素数      1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,      20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,      554400  };    int fact[]={//反素数约数个数      1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,      144,160,168,180,192,200,216  };  struct {int num;char name[11];}child[500005];void build(int l,int r,int rt){if(l==r){tree[rt]=1;return ;}tree[rt]=r-l+1;int m=(l+r)>>1;build(lson);build(rson);}int del(int pos,int l,int r,int rt){int ans;tree[rt]--;if(l==r)return l;int m=(l+r)>>1;if(pos<=tree[rt<<1])ans=del(pos,lson);elseans=del(pos-tree[rt<<1],rson);return ans;}int main(){int i,n,k,mod,p,flag,t,pre,tmp;bool circle;//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);while(scanf("%d%d",&n,&k)!=-1){tmp=k;flag=0;mod=n;circle=false;build(1,n,1);for(i=1;i<=n;i++)scanf("%s%d",&child[i].name,&child[i].num);for(i=0;RPrime[i]<=n;i++){flag=RPrime[i];t=i;}child[0].num=0;p=pre=0;for(i=1;i<=flag;i++){if(child[p].num>0)    k=((k+child[p].num-2)%mod+mod)%mod+1;else    k=((k+child[p].num-1)%mod+mod)%mod+1;p=del(k,1,n,1);if(p==pre){circle=true;break;}pre=p;mod--;}if(circle)printf("%s %d\n",child[tmp].name,1);elseprintf("%s %d\n",child[p].name,fact[t]);}return 0;}

另外介绍下反素数以及反素数的求法

http://blog.csdn.net/ACdreamers/article/details/25049767

反素数的求法
http://blog.csdn.net/hnust_xiehonghao/article/details/8873218