cf 153.div2 D. Playing with Permutations

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D. Playing with Permutations

比赛的时候一直没看懂题，后来看了下，其实很简单，就是给你一个变换序列q ，有两种变换1和2，1就是令Pi=Pqi，2则相反，互为逆变换，所以只要单独求出每种变换要多少少次，即可判断出来能否成功变换。

/*author:jxylang:C/C++university:China,Xidian University**If you need to reprint,please indicate the source***/#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>#include <map>#define INF 1E9using namespace std;int q[100];string aim="";char p[105];int i,n;string c1(string s){    for(i=0;i<n;i++)      p[i]=s[q[i]];    return p;}string c2(string s){    for(i=0;i<n;i++)      p[q[i]]=s[i];    return p;}string now="",nn;int k1,k2;int main(){    int k,t,lvl;    scanf("%d%d",&n,&k);    for(i=0;i<n;i++)    {        scanf("%d",&q[i]);        q[i]--;    }    for(i=0;i<n;i++)    {        scanf("%d",&t);        aim.push_back(t);        now.push_back(i+1);    }    nn=now;    for(k1=0;k1<=k&&now!=aim;now=c1(now),k1++);    for(k2=0;k2<=k&&nn!=aim;nn=c2(nn),k2++);    if((k-k2)%2&&(k-k1)%2||(k>1&&k1==1&&k2==1)||!k1)cout<<"NO"<<endl;    else cout<<"YES"<<endl;}