POJ 2886 Who Gets the Most Candies?

思路：
我脑子里总把他想成一个环，其实线段+取模就是环。
利用线段树来寻找第k个空位的位置，第一个问题就解决了。
第二个问题是约瑟夫环左右走的问题，确实也翻了好多题解，大家好像都顺其自然就写出了算式，
果然还是太菜了，我理解了好久才有点明白为啥正向要减一。
每次跳出一个人m时，在他之后的数rank-1，这也就是为何正向走要减一的原因。
那走远一些，走到m后面没改变的数rank不用减了吧？
确实如此，要走到m后边，必定会经过取模操作，而取模操作的过程就进行了+1。
剩下的问题就只剩下因子最大的数的预处理了。

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <functional>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define CLOSE() ios::sync_with_stdio(false)#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define Mid(l, r) (l + r) >> 1#define FOR(i, a, b) for(int i = a; i < b; ++i)#define LL long long#define maxn 500005#define maxm 1000005#define mod  1000000007#define INF 10000000#define EPS 1e-6#define N 12using namespace std;//-------------------------CHC------------------------------//#define lson cur<<1, l, mid#define rson cur<<1|1, mid+1, rchar name[maxn][N];int k[maxn], sum[maxn<<2];int n, m;const int rp[] = { 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 };const int fac[] = { 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 };void PushUp(int cur) {sum[cur] = sum[cur << 1] + sum[cur << 1 | 1];}void build(int cur, int l, int r) {if (l == r) sum[cur] = 1;else {int mid = Mid(l, r);build(lson);build(rson);PushUp(cur);}}int update(int i, int cur, int l, int r) {if (l == r) { sum[cur] = 0; return l; }int mid = Mid(l, r);int ret = 0;if (i <= sum[cur << 1]) ret = update(i, lson);else ret = update(i - sum[cur << 1], rson);PushUp(cur);return ret;}int pre() {int ret = 0;while (rp[ret] <= n) ret++;return ret-1;}int main() {while (~scanf("%d%d", &n, &m)) {FOR(i, 1, n + 1) scanf("%s%d", name[i], &k[i]);int idx = pre();build(1, 1, n);int num, ans;FOR(i, 1, rp[idx]) {ans = update(m, 1, 1, n);num = sum[1];if(k[ans] > 0) m = ((m - 2 + k[ans]) % num + num) % num + 1;else m = ((m - 1 + k[ans]) % num + num) % num + 1;}ans = update(m, 1, 1, n);printf("%s %d\n", name[ans], fac[idx]);}return 0;}