[Uoj 74] "UR #6" crack password

Title Link: UOJ-74

Problem analysis

Topic, the first character of the string S is moved to the end, and the other characters move forward one position, and f (s) changes from Hi to hi+1.

Let's analyze this process: Suppose the first character is C, (hi-26^ (n-1) * C) * + + c = hi+1

* hi-hi+1 = (26^n-1) * C

c = (* hi-hi+1) * INV (26^n-1)

Then each C can be directly solved, because the topic guaranteed to have the solution, so each C solution out of must be [0, 25] number.

Well.. It looks very right. Submit up .... wa.50

Why do we have to have 50 points? because: " when there is division, be sure to consider the case of no inverse element!" "

When (26^n-1)% P = 0, the 26^n-1 is not inverse. This point has 5 "Qaq" in the data.

So in the preceding equation, hi-hi+1 = (26^n-1) * C has a term coefficient of 0, which is eliminated and becomes Hi * = hi+1.

Because the equation is not related to C, C is any one letter that can be set up.

Then we have to find this S from f (s) = Hi, we know that f (s) is equivalent to the string as a 26 binary number, then here's Hi * = hi+1.

So we think of H0 as a 26 binary number to find the corresponding string, then move left one is equivalent to the H by 26, still conforms to f (S) = hi+1.

Code

#include <iostream> #include <cstdlib> #include <cstdio> #include <cmath> #include <cstring > #include <algorithm>using namespace std;const int maxn = 100000 + 5;typedef long long ll;int N, p, T;int H[MAXN] ; LL A0, Temp; LL X[MAXN]; ll Pow (ll a, int b) {LL ret = 1, f = a;while (b) {if (b & 1) {ret *= f;ret%= p;} b >>= 1;f *= f;f%= p;} return ret;} ll NY (ll x) {x = ((x% p) + p)% P;return Pow (x, p-2);} int main () {scanf ("%d%d", &n, &p), for (int i = 0; i < n; ++i) scanf ("%d", &h[i]); if (Pow (26LL, n)! = 1) {Tem p = NY (Pow (26LL, N)-1), for (int i = 0; i < n; ++i) {if (i = = n-1) T = 0;else T = i + 1; X[i] = ((LL) h[i] *-(LL) h[t])% p * TEMP; X[i] = ((X[i]% p) + p)% p;}} Else{int Pos = N;while (H[0]) {X[--pos] = h[0]% 26; H[0]/= 26;}} for (int i = 0; i < n; ++i) printf ("%c", ' a ' + x[i]);p rintf ("\ n"); return 0;}

　　

[Uoj 74] "UR #6" crack password