POJ-2886 Who Gets the Most Candies?

Who Gets the Most Candies?

Time Limit: 5000MS Memory Limit: 131072KCase 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

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前言：胡大大给出的练习题。吗，没办法搜别人的题解了。然后努力改写成了胡大大的姿势。

思路：首先分解成若干个子问题：

1. 糖果的个数（第n个出圈的人获得的糖果个数 等同于n的约数的个数）

2. 孩子出圈的过程是一个约瑟夫环游戏，但是初始孩子是第K个，并且由于间隔人数不一定，所以必须模拟

3. 由于数据范围和时限，采用线段树存储孩子们，区间值是孩子数。根据孩子数查询孩子的初始编号

对于问题1：反素数表，暴力打出来即可，但是姿势要对，根据“n = i * j, F[n]有两个约数”的方式暴力。

对于问题2：数下一个孩子的时候一定要细心，从第k个孩子出圈开始，数C[i]个孩子，下一个出圈的孩子是剩下孩子当中他的第几个孩子，（数孩子花了我一个小时，数死早没办法）

对于问题3：由于每次都得到下一个出圈的是第几个孩子，在线段树中利用树状数组的思想去查询。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <iostream>#define INF 0x3f3f3f3f#define LL long longusing namespace std;/****************************************/#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1const int N = 5e5+5;char name[N][15];int card[N], F[N], tree[N<<2];void push_up(int rt){tree[rt] = tree[rt<<1] + tree[rt<<1|1];}void build(int l, int r, int rt){if(l == r) {tree[rt] = 1;//记录的是区间孩子数return ;}int m = (l+r)>>1;build(lson); build(rson);push_up(rt);}int query(int pos, int l, int r, int rt){if(l == r) {tree[rt]--;return l;}int ret, m = (l+r)>>1;//利用树状数组的思想查询if(pos <= tree[rt<<1]) ret = query(pos, lson);else ret = query(pos - tree[rt<<1], rson);push_up(rt);return ret;}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endiffor(int i = 1; i*i <= N;  i++) {int j;for(j = i+1; j*i <= N; j++) {F[i*j] += 2;}F[i*i]++;}//暴力打出反素数表int n, k;while(~scanf("%d%d", &n, &k)) {for(int i = 0; i < n; i++) {scanf("%s%d", name[i], &card[i]);}build(0, n-1, 1);int id, maxC = 0, ans;for(int p = 1; p <= n; p++) {//到树中所有孩子出圈为止id = query(k, 0, n-1, 1);//根据k找到这个孩子的idif(maxC < F[p]) {maxC = F[p];ans = id;}if(p == n) break;//接下来是约瑟夫环int rest = n-p;//剩余孩子数k--;//因为取余运算，从0开始给孩子编号if(card[id] > 0) {//顺时针数//由于第k个孩子出圈，所以从第k-1个孩子开始数k = (k - 1 + card[id]) % rest+1;}else {//逆时针数//由于第k个孩子已经出圈，所以他左边的孩子成为第k个孩子k = ((k + card[id]) % rest + rest) % rest+1;}}printf("%s %d\n", name[ans], maxC);}return 0;}