POJ Who Gets the Most Candies? 2886(线段树)

Who Gets the Most Candies?

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 13876 Accepted: 4385Case 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

题意：(copy)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；

分析：第一印象是约瑟夫环，可是n太大了，用线段树找出第几个出来的编号，首先要找出第几个出来最大

#include <stdio.h>#include <string.h>#include <algorithm>#include <stdlib.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define maxx 510000struct{    char name[15];    int h;}q[maxx];struct node{    int l,r,sum;//sum代表区间还剩多少人}tree[maxx<<2];int v[maxx],id,ma;void build(int l,int r,int rt){    tree[rt].sum=(r-l+1);    tree[rt].l=l;    tree[rt].r=r;    if(l==r)        return ;    int m=(l+r)>>1;    build(lson);    build(rson);}int gengxin(int k,int rt){    tree[rt].sum--;   //每删除一个就更新区间    if(tree[rt].l==tree[rt].r)        return tree[rt].l;    if(tree[rt<<1].sum>=k)//如果左孩子剩的人数大于ｋ        return gengxin(k,rt<<1);    else        return gengxin(k-tree[rt<<1].sum,rt<<1|1);}void zhao(int n)//找第几个最大{    int i,j;    memset(v,0,sizeof(v));    for(i=1;i<=n;i++)    {        v[i]++;        for(j=2*i;j<=n;j+=i)            v[j]++;    }    ma=1;    for(i=2;i<=n;i++)    {        if(v[i]>ma)        {            ma=v[i];            id=i;        }    }}int main(){    int n,m,i,j,k;    while(~scanf("%d%d",&n,&k))    {        zhao(n);        for(i=1;i<=n;i++)        {            scanf("%s%d",q[i].name,&q[i].h);        }        build(1,n,1);        m=id-1;//找前id个就行        int pos=gengxin(k,1);//先找出来第一个        int mod=tree[1].sum;        while(m--)        {            if(q[pos].h>0)//k代表删除后这个人排第几            {               k=(k-1+q[pos].h-1+mod)%mod+1;//k-1是防止k%mod==0的情况，而q[pos].h-1是　当这个本身删掉，            }                                                                 //他的下一个，就是第k个，只需要在这一个的基础上移动q[pos].h-1个            else            {                k=((k-1+q[pos].h)%mod+mod)%mod+1;            }            pos=gengxin(k,1);            mod=tree[1].sum;        }        //printf("%d\n",pos);        printf("%s %d\n",q[pos].name,ma);    }    return 0;}