约瑟夫环用线段数搞，一脸搞不出来的样子。反素数，太神了，先打表，然后就可以 O(1)找到因子数最多的。ps：哎。这题也是看着题解撸的。
#include <cstdio> #include <cstring> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> #include<math.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 int k; int sum[2222222]; int chart[35][2] = { 498960, 200, 332640, 192, 277200, 180, 221760, 168, 166320, 160, 110880, 144, 83160, 128, 55440, 120, 50400, 108, 45360, 100, 27720, 96, 25200, 90, 20160, 84, 15120, 80, 10080, 72, 7560, 64, 5040, 60, 2520, 48, 1680, 40, 1260, 36, 840, 32, 720, 30, 360, 24, 240, 20, 180, 18, 120, 16, 60, 12, 48, 10, 36, 9, 24, 8, 12, 6, 6, 4, 4, 3, 2, 2, 1, 1 }; void up(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; } void build(int l, int r, int rt) { if (l == r){ sum[rt] = 1; return; } int mid = (l + r) >> 1; build(lson); build(rson); up(rt); } void update(int key, int l, int r, int rt) { if (l == r){ sum[rt] = 0; k = l; return; } int mid = (l + r) >> 1; if (key <= sum[rt << 1]) update(key, lson); else update(key - sum[rt << 1], rson); up(rt); } int ask(int L, int R, int l, int r, int rt) { if (L <= l&&r <= R) return sum[rt]; int ans = 0; int mid = (l + r) >> 1; if (L <= mid) ans += ask(L, R, lson); if (R>mid) ans += ask(L, R, rson); return ans; } char str[555555][11]; int a[555555]; int main() { int n; while (scanf("%d%d", &n, &k) != EOF){ int cnt = 0; while (n<chart[cnt][0]) cnt++; int t = chart[cnt][0]; for (int i = 1; i <= n; i++){ scanf("%s%d", str[i], &a[i]); } build(1, n, 1); int m = n; int now = k; for (int i = 0; i<t - 1; i++){ update(now, 1, n, 1); m--; if (a[k] % m == 0){ if (a[k]>0) a[k] = m; else a[k] = 1; } else{ a[k] %= m; if (a[k]<0) a[k] += m + 1; } int cc = ask(1, k, 1, n, 1); int tt = m - cc; if (a[k] <= tt) now = a[k] + cc; else now = a[k] - tt; } update(now, 1, n, 1);//m 最后为0 printf("%s %d ", str[k], chart[cnt][1]); } return 0; }