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  • URAL 2037 Richness of binary words （回文子串，找规律）

    Richness of binary words

    题目链接：

    http://acm.hust.edu.cn/vjudge/contest/126823#problem/B

    Description

    For each integer i from 1 to n, you must print a string s i of length n consisting of letters ‘a’ and ‘b’ only. The string s i must contain exactly i distinct 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 : NO
    4 : aaaa


    ##题意： 要求输出n个长度为n的字符串(只能放a或b)： 要求Si中不同回文子串的个数恰为i.
    ##题解： 本质就是找出一种构造方式使得回文子串个数不会再增加. 参考2015-ICPC合肥现场赛H题： http://blog.csdn.net/snowy_smile/article/details/49870109 http://blog.csdn.net/keshuai19940722/article/details/49839359
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 210000 #define mod 100000007 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

    int n;

    int main(int argc, char const *argv[])
    {
    //IN;

    while(scanf("%d", &n) != EOF)
{
    if(n == 1) {
        printf("1 : a
");
        continue;
    }
    if(n == 2) {
        printf("1 : NO
");
        printf("2 : ab
");
        continue;
    }

    if(n < 8) {
        for(int i=1; i<n; i++)
            printf("%d : NO
", i);
        printf("%d : ", n);
        for(int i=1; i<=n; i++)
            putchar('a');
        printf("
");
        continue;
    }

    if(n == 8) {
        for(int i=1; i<7; i++)
            printf("%d : NO
", i);
        printf("7 : aababbaa
");
        printf("8 : ");
        for(int i=1; i<=n; i++)
            putchar('a');
        printf("
");
        continue;
    }

    char str[] = "abaabb";
    for(int i=1; i<=7; i++)
        printf("%d : NO
", i);
    for(int i=8; i<=n; i++) {
        printf("%d : ", i);
        for(int j=1,cur=0; j<=n; j++) {
            if(j <= i-8) putchar('a');
            else {
                if(cur == 6) cur = 0;
                putchar(str[cur++]);
            }
        }
        printf("
");
    }
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5746764.html
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