————————————————————————————————————————题解
唯一没有想出来的题
我们发现以上两种操作
a0,a3,a6,a9……的相对位置不变
a1,a4,a7,a10……的相对位置不变
a2,a5,a8,a12……的相对位置不变
所以只要记下a0,a1,a2最后的位置，就可以推出剩下的数的位置
1 #include <iostream> 2 #include <queue> 3 #include <set> 4 #include <cstdio> 5 #include <cstring> 6 #include <vector> 7 #include <algorithm> 8 #define siji(i,x,y) for(int i=x;i<=y;++i) 9 #define gongzi(j,x,y) for(int j=x;j>=y;--j) 10 #define xiaosiji(i,x,y) for(int i=x;i<y;++i) 11 #define sigongzi(j,x,y) for(int j=x;j>y;--j) 12 #define ivorysi 13 #define inf 0x3f3f3f3f 14 #define mo 97797977 15 #define ha 974711 16 #define ba 47 17 #define fi first 18 #define se second 19 #define pii pair<int,int> 20 typedef long long ll; 21 using namespace std; 22 int n,q; 23 int se[300005],p,a; 24 int c[300005]; 25 int w[3]; 26 void solve() { 27 scanf("%d%d",&n,&q); 28 siji(i,0,3*n-1) scanf("%d",&se[i]); 29 w[0]=0;w[1]=1;w[2]=2; 30 siji(i,1,q) { 31 scanf("%d",&p); 32 if(p==1) { 33 scanf("%d",&a); 34 siji(i,0,2) { 35 w[i]=(w[i]+a+3*n)%(3*n); 36 } 37 } 38 else { 39 siji(i,0,2) { 40 if(w[i]%3==0) w[i]=(w[i]+2+3*n)%(3*n); 41 else if(w[i]%3==2) w[i]=(w[i]-2+3*n)%(3*n); 42 } 43 } 44 } 45 c[w[0]]=0;c[w[1]]=1;c[w[2]]=2; 46 47 xiaosiji(i,1,n) { 48 siji(j,0,2) { 49 c[(w[j]+i*3)%(3*n)]=i*3+j; 50 } 51 } 52 siji(i,0,3*n-1) { 53 printf("%d%c",se[c[i]]," "[i==3*n-1]); 54 } 55 } 56 int main(int argc, char const *argv[]) 57 { 58 #ifdef ivorysi 59 freopen("a.in","r",stdin); 60 freopen("a.ou","w",stdout); 61 #else 62 freopen("f1.in","r",stdin); 63 #endif 64 solve(); 65 return 0; 66 }