poj 2886

Who Gets the Most Candies? Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 14206 Accepted: 4491Case 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 2 Tom 2 Jack 4 Mary -1 Sam 1

Sample Output

Sam 3

线段树问题，用线段树维护一个动态线性序列，至于p【i】函数，则是求反素数，打表处理。

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <stack> #include <map> #include <set> #include <vector> #include <queue> #define mem(p,k) memset(p,k,sizeof(p)); #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define inf 0x6fffffff #define LL long long using namespace std; char s[500010][11]; int m[500010],sum[500010<<2],n,k; int pos[500010]; void build(int l,int r,int rt){ sum[rt]=r-l+1; if(l==r){ scanf("%s%d",s[l],m+l); return; } int m=(l+r)>>1; build(lson); build(rson); } void Rprime(){ mem(pos,0); int n=500000; int nn=(int)sqrt(n*1.0);//cout<<nn<<endl; for(int i=1;i<=nn;i++){ for(int j=i+1;j*i<=n;j++){ pos[i*j]+=2; } pos[i*i]++; } } int update(int k,int l,int r,int rt){ if(l<=r)sum[rt]--; if(l>=r){ return l; } int m=(l+r)>>1; if(k<=sum[rt<<1])return update(k,lson); else return update(k-sum[rt<<1],rson); } int main() { Rprime(); while(cin>>n>>k){ build(1,n,1); int t=0,maxx=0,maxk; while(t++<n){ int p=update(k,1,n,1); if(pos[t]>pos[maxx]){ maxx=t; maxk=p; } int mod=n-t;//cout<<p<<endl; if(m[p]>0){ k--; } if(mod)k=((k-1+m[p])%mod+mod)%mod+1; }//cout<<"asd"<<endl; printf("%s %d\n",s[maxk],pos[maxx]); } return 0; }