POJ 2886 Who Gets the Most Candies?(線段樹+反素數)

畫船聽雨發表於2014-02-22

這個反素數不太會，這道題目其實也不是自己想出來的看了別人的部落格：http://blog.csdn.net/weiguang_123/article/details/7880875

裡面說的很詳細，我就不在多說了啊。

Who Gets the Most Candies?
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 8786   Accepted: 2675
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
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 5000000;
using namespace std;

struct node
{
    int sum;
    int l, r;
} f[maxn*4];

char name[maxn][20];
int vis[maxn];

int RPrime[]=
{
    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
};  //反素數

int fact[]=
{
    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
};  //反素數約數個數

void Bulid(int l, int r, int site)
{
    f[site].l = l;
    f[site].r = r;
    f[site].sum = r-l+1;
    if(l == r)
        return ;
    int mid = (l+r)>>1;
    Bulid(l, mid, site<<1);
    Bulid(mid+1, r, site<<1|1);
}

int Search(int site, int num)
{
    f[site].sum--;
    if(f[site].l == f[site].r)
        return f[site].l;
    if(f[site<<1].sum >= num)
        return Search(site<<1, num);
    else
        return Search(site<<1|1, num-f[site<<1].sum);
}

int main()
{
    int n, k;
    while(scanf("%d %d",&n, &k) != EOF)
    {
        Bulid(1, n, 1);
        int p;
        int _max = 0;
        for(int i = 1; i <= n; i++)
            scanf("%s %d",name[i], &vis[i]);
        int i = 0;
        while(1)
        {
            if(RPrime[i] > n)
            {
                p = RPrime[i-1];
                _max = fact[i-1];
                break;
            }
            i++;
        }
        int t;
        for(i = 0; i < p; i++)
        {
            n--;
            t = Search(1,k);
            if(!n)
                break;
            if(vis[t] > 0)
                k=(k-1+vis[t]-1)%n+1;
            else
                k=((k-1+vis[t])%n+n)%n+1;
        }
        cout<<name[t]<<" "<<_max<<endl;
    }
    return 0;
}


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