(线段树,反素数)poj2886-Who Gets the Most Candies?
Posted 惜取少年时
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了(线段树,反素数)poj2886-Who Gets the Most Candies?相关的知识,希望对你有一定的参考价值。
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
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
线段树的每个结点表示代表的区间中有多少个人还在。分别记每次该出去的人的相对位置(即剩下的x个人1——x编号,第z号出去),以及这个人的绝对位置。
1 #include <iostream> 2 //#include<bits/stdc++.h> 3 #include <stack> 4 #include <queue> 5 #include <cstdio> 6 #include <cstring> 7 #include <algorithm> 8 using namespace std; 9 typedef long long ll; 10 typedef unsigned long long ull; 11 const int MAX=5e5+5; 12 const int antiprime[]={1,2,4,6,12,24,36,48,60,120,180,240,360,720,840, 13 1260,1680,2520,5040,7560,10080,15120,20160,25200, 14 27720,45360,50400,55440,83160,110880,166320,221760, 15 277200,332640,498960,554400,665280 16 }; 17 18 const int factorNum[]={1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60, 19 64,72,80,84,90,96,100,108,120,128,144,160,168,180, 20 192,200,216,224 21 }; 22 struct node 23 { 24 char name[10]; 25 int num; 26 }ren[MAX]; 27 int st[10*MAX]; 28 void init(int l,int r,int k) 29 { 30 st[k]=r-l; 31 if(l+1==r) 32 return; 33 init(l,(l+r)/2,2*k); 34 init((l+r)/2,r,2*k+1); 35 } 36 void push(int k) 37 { 38 if(k<=0) 39 return; 40 st[k]--; 41 push(k/2); 42 } 43 int update(int dis,int l,int r,int k) 44 { 45 if(l+1==r) 46 { 47 push(k); 48 return l; 49 } 50 if(dis<=st[2*k]) 51 return update(dis,l,(l+r)/2,2*k); 52 else 53 return update(dis-st[2*k],(l+r)/2,r,2*k+1); 54 } 55 int main() 56 { 57 int n,k;//n为初始人数,k为每次该出去的人的相对序号(即此时剩的人编上1——x时的序号) 58 int cnt; 59 int i; 60 int pos;//该出去的人的真实序号(初始时这个人的编号) 61 while(~scanf("%d %d",&n,&k)) 62 { 63 init(1,n+1,1); 64 for(i=1;i<=n;i++) 65 { 66 scanf("%s %d",ren[i].name,&ren[i].num); 67 } 68 cnt=0; 69 while(n>=antiprime[cnt]) 70 cnt++; 71 cnt--; 72 pos=0; 73 ren[0].num=0; 74 for(i=1;i<=antiprime[cnt];i++) 75 { 76 if(ren[pos].num>0) 77 k=((k+ren[pos].num-2)%st[1]+st[1])%st[1]+1;//此时st[1]个人,k值为1——st[1],不能为0,故进行此操作 78 else 79 k=((k+ren[pos].num-1)%st[1]+st[1])%st[1]+1; 80 pos=update(k,1,n+1,1); 81 } 82 printf("%s %d\n",ren[pos].name,factorNum[cnt]); 83 } 84 }
以上是关于(线段树,反素数)poj2886-Who Gets the Most Candies?的主要内容,如果未能解决你的问题,请参考以下文章
POJ2886Who Gets the Most Candies?(线段树之约瑟夫)
poj2886 Who Gets the Most Candies?
[poj 2886] Who Gets the Most Candies? 线段树
POJ 2886 Who Gets the Most Candies? (线段树)