解法：
对于第二个串，循环移动能得到的字典序最小的串，可以直接用最小表示法搞定。
然后用最小表示的第二个串和第一个串做两次扩展KMP，一次正常求，另外一次将两个串都反转一下，然后扫一遍ex[]数组
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int N = (int)2e6+10; 6 char a[N],b[N],c[N]; 7 int next[N],exa[N],exb[N]; 8 void getnext(char *s){ 9 int len = strlen(s),cur = 0; 10 next[0] = len; 11 while(cur < len&&s[cur]==s[cur+1])cur++; 12 next[1] = cur;cur = 1; 13 for(int k = 2;k<len;k++){ 14 int p = cur + next[cur] - 1,L = next[k-cur]; 15 if(k + L -1 >= p){ 16 int j = (p-k+1)>0?(p-k+1):0; 17 while(k+j<len&&s[k+j]==s[j])j++; 18 next[k] = j; 19 cur = k; 20 }else 21 next[k] = L; 22 } 23 } 24 void exkmp(char *s1,char *s2,int *ex){ 25 getnext(s2); 26 int l1 = strlen(s1),l2 = strlen(s2),cur = 0; 27 while(cur < min(l1,l2)&&s1[cur]==s2[cur])cur++; 28 ex[0] = cur;cur = 0; 29 for(int k = 1;k < l1;k++){ 30 int p = cur + ex[cur] - 1,L = next[k-cur]; 31 if(k + L -1 >= p){ 32 int j = (p-k+1)>0?(p-k+1):0; 33 while(k+j<l1&&j<l2&&s1[k+j]==s2[j])j++; 34 ex[k] = j; 35 cur = k; 36 }else 37 ex[k] = L; 38 } 39 } 40 int MinRep(char *s){ 41 int i = 0,j = 1,k = 0,t,len = strlen(s); 42 while(i<len&&j<len&&k<len){ 43 t = s[(i+k)%len] - s[(j+k)%len]; 44 if(t==0)k++; 45 else{ 46 if(t>0) 47 i += k + 1; 48 else 49 j += k + 1; 50 if(i==j)j++; 51 k = 0; 52 } 53 } 54 return min(i,j); 55 } 56 inline int dis(int now,int sz){ 57 return min(now,sz-now-1); 58 } 59 int main(){ 60 int n,m; 61 while(~scanf("%d%d",&n,&m)){ 62 scanf("%s%s",a,b); 63 int start = MinRep(b); 64 for(int i = 0,j = start;i < m;i++,j = (j+1)%m)c[i] = b[j];c[m] = 0; 65 for(int i = n;i < n+n;i++)a[i] = a[i-n];a[n+n] = 0; 66 exkmp(a,c,exa); 67 reverse(a,a+2*n);reverse(c,c+m); 68 exkmp(a,c,exb); 69 reverse(exb,exb+n*2); 70 reverse(c,c+m); 71 reverse(a,a+2*n); 72 int ans = -1; 73 for(int i = 0;i < n;i++) 74 if(exa[i] == m || exa[i] + exb[i+m-1] == m-1){ 75 if(ans==-1)ans = i; 76 else if(dis(i,n)<dis(ans,n))ans = i; 77 } 78 puts(c); 79 for(int i = 0,j = ans;i < n;i++,j = (j+1)%n)putchar(a[j]);puts(""); 80 } 81 return 0; 82 }