TTTTTTTTTTTTTTTTTTT UVA 2045 Richness of words

J - Richness of words

Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice URAL 2045

Description

For each integer i from 1 to n, you must print a string s _i of length n consisting of lowercase Latin letters. The string s _i must contain exactly idistinct palindrome substrings. Two substrings are considered distinct if they are different as strings.

Input

The input contains one integer n (1 ≤ n ≤ 2000).

Output

You must print n lines. If for some i, the answer exists, print it in the form “ i :  s _i” where s _i is one of possible strings. Otherwise, print “i : NO”.

Sample Input

input	output
4	1 : NO 2 : NO 3 : abca 4 : bbca

题意：给你一个数字n（n<=2000），for(int i=1;i<=n;i++)，如果存在一个长为n的字符串，且其中有i个回文串，则输出

该字符串，否则输出NO；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef  long long  ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e6+10;

char s[2005][2005];
int flag[2005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        MM(s,'');
        if(n==1) {printf("1 : a
");continue;}
        if(n==2) {
                printf("1 : NO
");
                printf("2 : ab
");
                continue;
       }

       for(int i=1;i<=3;i++) s[n][i]='a'+i-1;
       for(int i=4;i<=n;i++) s[n][i]='z';
       int p=3;
       for(int i=n-1;i>=1;i--)
       {
           int j;
           for(j=1;j<=p;j++)
               s[i][j]=s[i+1][j];
           if(s[i][p]=='a') s[i][j]='b';
           if(s[i][p]=='b') s[i][j]='c';
           if(s[i][p]=='c') s[i][j]='a';
           for(j++;j<=n;j++) s[i][j]='z';
           p++;
       }
       for(int i=1;i<=2;i++) printf("%d : NO
",i);
       for(int i=3;i<=n;i++) printf("%d : %s
",i,s[i]+1);
    }
    return 0;
}

　　分析：刚开始想的是比如n==30，，那么首先就是

aaaaabcdefg...xyz，

aaaaabcaefg....zyz

aaaaabcaafg....xyz;

这样下去，，但是其实这样不是最优的；

正确做法：考虑abc三个字符，无论怎样循环都是不会有回文出现的，比如abcabcabc....

那么对于n==30的情况；

abczzzzzz...z;//30

abcazzzzz...z;//29

abcabzzzz...z;//28

abcabczzz...z;//27

abcabcazz...z;//26

所以这样下去，每次都能保证这一个比上面一个减1，然后特判下2,3；