POJ 3087 Shuffle ' m up linear congruence, violence difficulty: 2

http://poj.org/problem?id=3087

Set: S1={a1,a2,a3,... AC}

S2={ac+1,ac+2,ac+3,.... A2C}

The

Together to become

Ac+1,a1,ac+2,a2 ... A2c,ac

After one conversion, it becomes

S1={ac+1,a1,ac+2 ...}

s2={... A2C,AC}

Corresponds to the ordinal of each number that occurs before the change is

+1,+2,+3....-c,-c+1,.....

To think of the whole chain as a ring is also equivalent to:

+1,+2,+3....+c,+c+1,.......

A1, for example, must return to A1 after the a1->a2->a4->a7....c times.

So the whole string passes a certain number of transformations will be back to the original state, just to determine whether to return to the original state before the target State can be

#include <cstdio>//a= (a+c+1)% (2*c) #include <cstring>using namespace Std;const int Maxn=1002;int C;char s1[ Maxn],s2[maxn],aim[maxn],org[maxn],tmp[maxn];void Shuffle () {for (int i=0;i<c;i++) {Tmp[2*i]=s2[i]                ;        Tmp[2*i+1]=s1[i]; } tmp[2*c]=0;}        int main () {int T;        scanf ("%d", &t);                for (int ti=1;ti<=t;ti++) {scanf ("%d%s%s%s", &c,s1,s2,aim);                Shuffle ();                strcpy (ORG,TMP);                int Ans=1;                        if (strcmp (Tmp,aim) ==0) {printf ("%d 1\n", TI);                Continue                } bool Fl=false; while (strcmp (org,tmp)!=0| |                        Ans==1) {strncpy (s1,tmp,c);                        strncpy (S2,TMP+C,C);                        Shuffle ();                        ans++;                        if (strcmp (Tmp,aim) ==0) {fl=true;        printf ("%d%d\n", Ti,ans);                        Break        }} if (!FL) printf ("%d-1\n", TI); } return 0;}

　　

POJ 3087 Shuffle ' m up linear congruence, violence difficulty: 2