POJ3087Shuffle'm Up

洗牌，注意S1在下面
搜2n次足矣

洗牌：

S3i={S1(i+1)/2(i≡1(mod2))S2i/2(2|i)

分开就前面分S2，后面分S1

/*************************************************************************    > file name: d.cpp    > author: huhao    > mail: [e-mail]     > created time: 2017/5/13 14:55:39 ************************************************************************/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<string>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>using namespace std;#define inf 0x3f3f3f3f#define eps 1e-8#define rt return#define gc getchar()#define ll long long#define fr(i,a,b) for(int i=a,_end_=b;i<=_end_;i++)#define fd(i,a,b) for(int i=a,_end_=b;i>=_end_;i--)void swap(int a,int b){    a^=b^=a^=b;}int read(){    int r=1,t=0;    char c=gc;    while(c<'0'||c>'9')    {        if(c=='-')            r=-1;        else            r=1;        c=gc;    }    while(c>='0'&&c<='9')    {        t=(t<<1)+(t<<3)+(c^48);        c=gc;    }    rt r*t;}char readchar(int x,int y){    char c=gc;    while(c<x||c>y)        c=gc;    rt c;}double log(double x,double y){    rt log10(x)/log10(y);}int gcd(int x,int y){    rt y?gcd(y,x%y):x;}int lcm(int x,int y){    rt x*y/gcd(x,y);}char a[2010],b[2010],c[2010],d[2010];int n;bool check(){    fr(i,1,n<<1)        if(d[i]!=c[i])            rt 0;    rt 1;}int main(){#ifndef online_judge    freopen("", "r", stdin);    freopen("", "w", stdout);#endif    fr(interesting,1,read())    {        n=read();        fr(i,1,n)            a[i]=readchar('A','Z');        fr(i,1,n)            b[i]=readchar('A','Z');        fr(i,1,n<<1)            c[i]=readchar('A','Z');        fr(i,1,n)            d[i]=b[i];        fr(i,1,n)            d[i+n]=a[i];        if(check())        {            printf("%d %d\n",interesting,0);            goto end;        }        //puts(c+1);        //puts(d+1);        fr(ans,1,2*n)        {            fr(i,1,n)            {                d[(i<<1)-1]=b[i];                d[i<<1]=a[i];            }            //puts(d+1);            fr(i,1,n)                a[i]=d[i];            fr(i,1,n)                b[i]=d[n+i];            if(check())            {                printf("%d %d\n",interesting,ans);                goto end;            }        }        printf("%d -1\n",interesting);        end:;    }    rt 0;}