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

Who Gets the Most Candies?

Time Limit: 5000MS Memory Limit: 131072KB 64bit IO Format: %I64d & %I64u

Submit Status

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 2Tom 2Jack 4Mary -1Sam 1

Sample Output

Sam 3

线段树，刚刚开始时把name和val都放在树里了，没必要而且要超时，比较难的地方在怎么处理出下次更新时的节点在第几个，要想一想，约数个数先处理出来。

#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<algorithm>using namespace std;const int MAXN = 500010;struct node{int l, r, sum;}tree[MAXN*3];int father[MAXN];char name[MAXN][15];int val[MAXN];void build(int l, int r, int i){tree[i].l = l; tree[i].r = r;if (l == r){father[l] = i;tree[i].sum = 1;return;}int m = (l + r) >> 1, ls = i << 1, rs = ls + 1;build(l, m, ls);build(m + 1, r, rs);tree[i].sum = tree[ls].sum + tree[rs].sum;}int update(int len, int i){tree[i].sum--;if (tree[i].l == tree[i].r){tree[i].sum = 0;return tree[i].l;}int m = (tree[i].l + tree[i].r) >> 1, ls = i << 1, rs = ls + 1;if (tree[ls].sum >= len) return update(len, ls);else return update(len - tree[ls].sum, rs);}int yue[MAXN];int main(){int n, k;for (int i = 1; i <= 500000;i++)for (int j = 1; j*i <= 500000; j++)yue[i*j]++;while (scanf("%d%d", &n, &k) != EOF){build(1, n, 1);for (int i = 1; i <= n; i++){scanf("%s%d",name[i], &val[i]);}/*for (int i = 1; i <= n; i++){cout << i << " " << father[i] << " " << name[i] << " " << val[i] << endl;;}*/if (n == 1){cout << name[1] << " " <<1<< endl;continue;}int op = k;int cnt = n;char ans_name[15];int ans=0;for (int i = 1; i <= n; i++){int v=update(k, 1);cnt--;/*for (int i = 0; i < 15; i++)cout << i << " " << tree[i].sum << endl;            */if (yue[i] > ans){ans = yue[i];strcpy(ans_name, name[v]);}if (i == n) break;v = val[v];if (v>0){k = ((k-1+v-1)%cnt+cnt)%cnt+1;}else{k = ((k - 1 + v) % cnt + cnt) % cnt + 1;}}cout << ans_name << " " << ans << endl;}}