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poj 2886 Who Gets the Most Candies?(线段树+约瑟夫环+反素数)

2024-07-17 11:19:16   215人阅读

Who Gets the Most Candies?
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 9934 Accepted: 3050
Case 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


题解及代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <string>
#define maxn 500010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ALL %I64d
using namespace std;
typedef long long  ll;
int pos,ans;      //pos记录每次跳出的人的所处的叶子节点的位置，ans为最后输出最大的F(k)的值

const int antiprime[] = {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, 665280};
const int factorNum[] = {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, 224};


int find_max(int n,int &x)   //寻找1--n中最大的F(k)值，并返回k
{
    int i=0;
    while(i<35)
    {
        if(antiprime[i]>n)
            break;
        i++;
    }
    x=factorNum[i-1];
    return antiprime[i-1];
}


struct segment     //定义线段树的节点
{
    int l,r;
    int value;
    char s[12];    //叶子节点中信息，非叶子节点无用
    int dir;       //读者也可以去掉这里，自行优化空间
} son[maxn*3];

void PushUp(int rt)
{
    son[rt].value=http://www.mamicode.com/son[rt<<1].value+son[rt<<1|1].value;>


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