Multiplication method + two hnu13547 Lily ' game

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Test instructions: Gives an original string A, and a permutation B. For each operation, the string A is transformed by permutation B, and the number of odd digits is multiplied by 2. Ask the original string a after many operations, can get c*2^d. If available, the minimum number of operations is output, otherwise the output-1

Idea: We analyze, first of all, for the problem of permutation transformation, of course, the first step is to transform the transformation into a graph model, that is, a side i->b[i]. We can see that we get a lot of rings.

The answers to the different rings are independent. We consider each ring as a line of thought. If A[i] transformation can become c*2^d, multiplied by x 2, then the position I according to the graph walk, just go through x points, the number of these points is odd.

We use the multiplication method to deal with starting from a point u, down the 1th point, the 2nd point, the 4th point ... The position of the 2^k point, and saves the number of odd numbered points on the path.

If there are y odd points on the whole ring, we need x, we'll use x%y to simplify, then we'll go to the second answer in the last lap.

The final total complexity is O (n (logn) ^2), should be able to do O (Nlogn), but I multiply the method to write a comparison rub, give a more Logn

But the trick point of the problem is too many, leading to the time of the game has been card-

1. The answer will explode int

2. If the number of odd points on the ring is 0, a special sentence, or divide 0, will be re

3. First remove the 2 in C and add it to D

4. In the x%y, but also to pay special attention to the division of the time, divide the number of steps may not walk the complete X/y rings, the number of steps may be less.

#include <map> #include <set> #include <cmath> #include <ctime> #include <stack> #include <queue> #include <cstdio> #include <cctype> #include <bitset> #include <string> #include <vector> #include <cstring> #include <iostream> #include <algorithm> #include <functional > #define FUCK (x) cout<< "[" <<x<< "]"; #define FIN freopen ("Input.txt", "R", stdin); #define FOUT Freopen ("Output.txt", "w+", stdout); using namespace Std;typedef long Long ll;typedef pair<int, int> pii;const int MX = 1e5 + 5;const LL INF = 0x3f3f3f3f;int n;int z[mx], a[mx], Nlen;int nxt[mx][20], Val[mx][20];int belong[mx], Clen[MX], CV AL[MX], Bcnt;bool vis[mx];void BFS (int u, int dfn) {queue<int> Q;    Q.push (U); while (! Q.empty ()) {u = Q.front ();        Q.pop (); clen[dfn]++;        Vis[u] = 1;        if (U & 1) cval[dfn]++;        Belong[u] = DFN;        Nxt[u][0] = Z[u];        Val[u][0] = Z[u]% 2; if (!vis[z[u]]) {Vis[z[u]] = 1;        Q.push (Z[u]);    }}}void Presolve () {bcnt = 0;    memset (Vis, 0, sizeof (VIS));            for (int i = 1; I <= n; i++) {if (!vis[i]) {++bcnt;            CLEN[BCNT] = cval[bcnt] = 0;        BFS (i, bcnt); }} for (int i = 1, i <= nlen; i++) {for (int j = 1; J <= N; j + +) {Nxt[j][i] = Nxt[nxt[j][i            -1]][I-1];        Val[j][i] = Val[j][i-1] + val[nxt[j][i-1]][i-1];    }}}int get (int m, int u) {int ret = 0;            for (int i = 0; I <= nlen; i++) {if (M >> I & 1) {ret + = Val[u][i];        U = nxt[u][i]; }} return ret;}    LL Solve (int u, int d) {int block = Belong[u];    if (d = = 0) return 0;    if (cval[block] = = 0) return-1;    int k = (d-1)/cval[block];    LL ret = (ll) k * Clen[block];    D-= k * Cval[block];    int L = 0, r = clen[block], m;        while (L <= r) {m = (L + r) >> 1; int temp = Get (M, u);        if (temp >= d) r = m-1;    else L = m + 1; } return ret + R + 1;}    int Getmax (int n) {for (int i = 1; I >= 0; i--) {if (n >> I &) return i;    }}int Main () {//fin;        while (~SCANF ("%d", &n)) {for (int i = 1; I <= n; i++) {scanf ("%d", &a[i]);        } for (int i = 1; I <= n; i++) {scanf ("%d", &z[i]);        } Nlen = Getmax (n);        Presolve ();        int C, D;        LL ans =-1;        scanf ("%d%d", &c, &d);        while (c% 2 = = 0) d++, c/= 2;            for (int i = 1; I <= n; i++) {if (A[i]% c! = 0) Continue;            A[i]/= C;            int s = 0;            While (A[i]% 2 = = 0) a[i]/= 2, s++;            if (a[i]! = 1 | | s > D) continue;            LL ret = Solve (I, D-s);                if (ret! =-1) {if (ans = =-1) ans = ret;            else ans = min (ans, ret);    }} printf ("%d\n", ans); }    return 0;} 

Multiplication method + two hnu13547 Lily ' game