poj 2886 Who Gets the Most Candies? （树状数组+二分+反素数）

Who Gets the Most Candies?

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 11597 Accepted: 3616Case 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 theK-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. LetA denote the integer. If A is positive, the next child will be theA-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, thep-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 integersN (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

知道了一种能把当前剩余人中某人的id 和原始id关联起来的的方法。树状数组。sum（n）=i；就是原始位置为n的人目前在残存环形中位置是i。知道了有个反素数表。通过他
第几个人出来时候糖果数最大。然后模拟出那个最大糖果数的人。

#include<iostream>#include<sstream>#include<algorithm>#include<cstdio>#include<string.h>#include<cctype>#include<string>#include<cmath>#include<vector>#include<stack>#include<queue>#include<map>#include<set>using namespace std;const int INF=500003;int N,K;int bit[INF-3];char name [INF-2][16];int integer[INF-2];int a[39]= {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,            55440,83160,110880,166320,221760,277200,332640,498960,500001           };int b[39]= {1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,144,160,168,180,192,200,1314521};int lowbit(int x){    return x&(-x);}void add (int x,int val){    ++x;    while(x<INF)    {        bit[x]+=val;        x+=lowbit(x);    }}int sum(int x){    int s = 0;    while(x>0)    {        s += bit[x];        x-=lowbit(x);    }    return s;}int binary_Search(int id){    int l,r,mid;    int flag=1;    l=0,r=N;    while(r-l>1)    {        mid = (l + r)>>1;        if(sum(mid)<=id)        {            l=mid;        }        else        {            r=mid;        }    }    return l;}int main(){    while(cin>>N>>K)    {        --K;        for(int i=0; i<N; i++)        {            scanf("%s %d",name[i],&integer[i]);            add(i,1);        }        int candy = -1,index;        int iu = 0, Max = 0, p = 0;        while (a[iu] <= N)            iu++;        p = a[iu-1];        Max = b[iu-1];        for(int i=1; i<p; i++)        {            add(K,-1);            int mod = N-i;            int id=sum(K)+integer[K]+(integer[K]>0?-1:0);            id=(id%mod+mod)%mod;            K=binary_Search(id);        }        printf("%s %d\n",name[K],Max);    }    return 0;}