The tree array is more elegant than the line tree... We have done a good job in anti-prime number tabulation, that is, the formula for pushing the next position from the current position has not been developed, so we can see other people's...

If (P [now]. x> 0) NXT = (NXT-1 + P [now]. X) % leave;

Else NXT = (NXT + P [now]. X) % leave + leave) % leave;

Then, in the tree array, find the number (Binary) in the NXT value )...

View code

// # Pragma comment (linker, "/Stack: 327680000,327680000 ") # include <iostream> # include <cstdio> # include <cmath> # include <vector> # include <cstring> # include <algorithm> # include <string> # include <set> # include <functional> # include <numeric> # include <sstream> # include <stack> # include <map> # include <queue> # define Cl (ARR, val) memset (ARR, Val, sizeof (ARR) # define rep (I, n) for (I) = 0; (I) <(N ); ++ (I) # define For (I, L, H) for (I) = (l); (I) <= (h); ++ (I) # define Ford (I, h, L) for (I) = (h); (I)> = (l); -- (I) # define L (x) (X) <1 # define R (x) <1 | 1 # define mid (L, R) (L + r)> 1 # define min (X, y) (x) <(y )? (X): (y) # define max (x, y) (x) <(y )? (Y): (x) # define e (x) (1 <(x) # define iabs (x) <0? -(X): (x) # define out (x) printf ("% i64d \ n", x) # define read () freopen ("data. in "," r ", stdin) # define write () freopen (" data. out "," W ", stdout); const double EPS = 1e-8; typedef long ll; const int INF = ~ 0u> 2; using namespace STD; const int n = 500010; const int M = 1024; struct node {int A, B; node () {} node (INT X, int y): A (x), B (y) {} bool operator <(const node CMP) const {return B <CMP. B ;}} f [m]; bool vis [m]; int prime [m]; int CNT, num; void get_prime () {int I, j; CL (VIS, true); for (I = 2; I <m; ++ I) {for (j = I * I; j <m; j + = I) {vis [J] = false ;}} CNT = 0; for (I = 2; I <m; ++ I) {If (vis [I]) Prime [CNT ++] = I ;}} void solve (INT pnum, int Val, int POs, int Lim) {If (Val> N) return; f [num ++] = node (pnum, Val); LL nlim, nval, npnum; nval = val; nlim = 0; npnum = 1; while (nlim <Lim & nval <= N) {nlim ++; npnum ++; nval * = prime [POS]; If (nval <= N) solve (npnum * pnum, nval, POS + 1, nlim) ;}} void get_antiprime () {get_prime (); num = 0; solve (1, 1, 0, 32); sort (F, F + num);} // --------- all of the above Table ------------------ int C [N], n; int lowbit (int I) {return I & (-I) ;} void add (INT P, int Val) {While (P <= N) {C [p] + = val; P + = lowbit (p) ;}} int sum (INT p) {int res = 0; while (P> 0) {res + = C [p]; P-= lowbit (p);} return res;} struct pnode {char name [20]; int X;} p [N]; int BS (int K) {int L = 1, R = N, mid, TMP; while (R-l> = 1) {mid = mid (L, R); TMP = sum (MID); If (TMP> = k) r = mid; else l = mi D + 1;} // printf ("% d \ n", l, R); return r;} int main () {// read (); get_antiprime (); int K, I, G, MX; int now, CNT, NXT, NN, M; while (scanf ("% d", & N, & K )! = EOF) {Cl (C, 0); for (I = 1; I <= N; ++ I) {scanf ("% S % d ", P [I]. name, & P [I]. x); add (I, 1);} MX = 0; G = 0; for (I = 0; I <num & F [I]. B <= N; ++ I) {If (F [I]. a> MX) {MX = f [I]. a; G = f [I]. B ;}} now = NXT = K; CNT = 1; while (CNT <G) {nn = N-CNT; M = P [now]. x; If (M> 0) NXT = (NXT-1 + M % NN) % nn; else NXT = (NXT + M) % Nn + nn) % nn; if (NXT = 0) NXT = nn; add (now,-1); // printf ("% d % s \ n", now, P [now]. name); now = BS (NXT); // printf ("% d \ n", now, NXT, NN); CNT ++ ;} printf ("% S % d \ n", P [now]. name, MX) ;}return 0 ;}