POJ 2886 Who Gets the Most Candies?【线段树单点更新+反素数打表】【好题】

Who Gets the Most Candies?

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 14886		Accepted: 4687
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

Source

POJ Monthly--2006.07.30, Sempr

题意：n个熊孩子按顺时针排列，每个人受伤都有一张牌，牌上有一个数字，从第k个孩子开始出队，出队的熊孩子卡上数字是K,则顺时针第k人是下一个出队的，负数则逆时针，第P个出队的会得到的糖果数是p的因子个数，输出得到最多糖果的人和他的糖果数，如果有多个，则输出最先出队的

线段树存储的是每个子树中还有几个人，因子数可以用反素数打表求得

反素数打表参考HDU2521

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn = 500000 + 10;
const int mod = 1e9 + 7;
const double pi = 3.14159265358979;
int s[40] = { 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[40] = { 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 };

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;

int n, k, pos, val[500111], ans;
char name[500111][20];

struct seg {
	int l, r, n;
}a[2000100];

void init(int l, int r, int rt)
{
	a[rt].l = l;
	a[rt].r = r;
	a[rt].n = r - l + 1;
	int m = (l + r) >> 1;
	if (l != r) {
		init(l, m, rt * 2);
		init(m + 1, r, rt * 2 + 1);
	}
}

int insert(int i, int x)
{
	a[i].n--;
	if (a[i].l == a[i].r)
	{
		return a[i].l;
	}
	if (x <= a[i * 2].n) insert(i * 2, x);
	else insert(i * 2 + 1, x - a[i * 2].n);
}

int main()
{
	while (~scanf("%d%d", &n, &k))
	{
		for (int i = 1; i <= n; i++)
		{
			scanf("%s%d", name[i], &val[i]);
		}
		init(1, n, 1);
		int  i = 0;
		while (s[i] <= n) i++;
		int p = s[i - 1];
		ans = b[i - 1];
		while (p--)
		{
			n--;
			pos = insert(1, k);
			if (!n) break;
			if (val[pos] >= 0)
			{
				k = (k - 1 + val[pos] - 1) % n + 1;
			}
			else
			{
				k = ((k - 1 + val[pos]) % n + n) % n + 1;
			}
		}
		printf("%s %d
", name[pos], ans);
	}

	return 0;
}