hdu 4731 Minimum palindrome 找规律 (2013 ACM/ICPC Asia Regional Chengdu Online 1004)
2013-09-14 16:56
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#include <cstdio> #include <string> #include <cmath> #include <iostream> #include <algorithm> using namespace std; #define LL __int64 int main() { int T,tt=0; cin>>T; while(T--) { int i,j,k,n,m; cin>>m>>n; cout<<"Case #"<<++tt<<": "; if(m==1){ for(i=0;i<n;i++) cout<<"a"; cout<<endl; } else if(m>2) { for(i=0;i<n/3;i++) cout<<"abc"; if(n%3==1)cout<<"a"<<endl; else if(n%3==2)cout<<"ab"<<endl; else cout<<endl; } else { if(n<=4) { for(i=0;i<(n+1)/2;i++) cout<<"a"; for(i=0;i<n/2;i++) cout<<"b"; cout<<endl; } else if(n==5)cout<<"aaaba"<<endl; else if(n==6)cout<<"aaabab"<<endl; else if(n==7)cout<<"aaababb"<<endl; else if(n==8)cout<<"aaababbb"<<endl; else { cout<<"aa"; n=n-2; for(i=0;i<n/6;i++) cout<<"aababb"; if(n%6==1)cout<<"a"<<endl; else if(n%6==2)cout<<"aa"<<endl; else if(n%6==3) cout<<"aaa"<<endl; else if(n%6==4) cout<<"aaaa"<<endl; else if(n%6==5) cout<<"aabab"<<endl; else cout<<endl; } } } return 0; } /* 题意,用前m个字母,组成长度为n的字符串,使得子串中回文数最小 当m=1时,必定全是a。 当m>2时,必定全是abc循环,因为要构成回文,回文中左右端点相同,且第二级相同。举例证明: a的右边必定是b,a的左边必定是c。所以含a的必定不能成回文。而不含a的只有bc,所以abc循环回文数最小。 当m=2时,思考用上述m>2的循环来接。当只有a,b两个字母。所以要两个字母组合,有aa,bb,ab,ba 任意取三个判断下,回文数最小,字典序最小。可以发现aababb循环的回文数最小,为4.当n<=8时需要特判, 因为n<=8时,最小回文数小于4.当n>=9时,开头添加aa不会增大回文数,且使字典序最小。删去aa后,接着就是 aababb循环了。 注意的是,aababb循环有剩余的时候,还需要特判。因为剩余的数不会有右边的数影响,可以用更小的字典序。 */
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