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,‘\0‘);
if(n==1) {printf("1 : a\n");continue;}
if(n==2) {
printf("1 : NO\n");
printf("2 : ab\n");
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\n",i);
for(int i=3;i<=n;i++) printf("%d : %s\n",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；