zoukankan      html  css  js  c++  java
  • Codeforces Round #339 (Div. 1) C. Necklace 构造题

    C. Necklace

    题目连接:

    http://www.codeforces.com/contest/613/problem/C

    Description

    Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).

    Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.

    Input

    The first line of the input contains a single number n (1 ≤ n ≤ 26) — the number of colors of beads. The second line contains after n positive integers ai — the quantity of beads of i-th color. It is guaranteed that the sum of ai is at least 2 and does not exceed 100 000.

    Output

    In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.

    Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.

    Sample Input

    3
    4 2 1

    Sample Output

    1
    abacaba

    Hint

    题意

    你有n个颜色的珠子,你要把所有珠子都穿成一个圆环

    然后你可以从圆环的某一个地方剪开,使得这个链是一个回文串

    然后问你怎么样去构造那个圆环,可以使得构成回文串的链最多

    题解

    考虑对称性,如果答案是ans的话,那么这ans刀一定是等分这个圆环的

    每一个块内的珠子数量肯定是相同的

    那么显然要满足这两个特性,ans = gcd(a[i])

    有两个特殊情况需要考虑

    如果存在大于1个颜色的珠子为奇数的话,答案为0,显然

    如果有一个颜色的珠子为奇数的话,只要把这个珠子放在中间就好了,相当于把其他的偶数都砍一半,然后去摆放。

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    int p[30];
    int odd = 0;
    int gcd(int a,int b)
    {
        if(b==0)return a;
        return gcd(b,a%b);
    }
    int main()
    {
        int n;scanf("%d",&n);
        int t = 0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&p[i]);
            if(p[i]%2==1)t=i,odd++;
        }
        if(odd>1)
        {
            printf("0
    ");
            for(int i=0;i<n;i++)
                for(int j=0;j<p[i];j++)
                    printf("%c",'a'+i);
            printf("
    ");
            return 0;
        }
        if(odd==1)
        {
            int ans = 0;
            for(int i=0;i<n;i++)
            {
                if(i==t)ans=gcd(p[i],ans);
                else ans=gcd(p[i]/2,ans);
            }
            printf("%d
    ",ans);
            for(int i=0;i<ans;i++)
            {
                for(int j=0;j<n;j++)
                    for(int k=0;k<p[j]/ans/2;k++)
                        printf("%c",'a'+j);
                printf("%c",'a'+t);
                for(int j=n-1;j>=0;j--)
                    for(int k=0;k<p[j]/ans/2;k++)
                        printf("%c",'a'+j);
            }
        }
        else
        {
            int ans = 0;
            for(int i=0;i<n;i++)
                ans=gcd(p[i],ans);
            printf("%d
    ",ans);
            for(int i=0;i<ans/2;i++)
            {
                for(int j=0;j<n;j++)
                    for(int k=0;k<p[j]/ans;k++)
                        printf("%c",'a'+j);
                for(int j=n-1;j>=0;j--)
                    for(int k=0;k<p[j]/ans;k++)
                        printf("%c",'a'+j);
            }
        }
    }
  • 相关阅读:
    写给想要做自动化测试的人(更新篇)
    testng 教程之使用参数的一些tricks配合使用reportng
    testng教程之testng.xml的配置和使用,以及参数传递
    testng 教程
    博客园计算园龄不正确,请管理员确认
    selenium 总结篇,常见方法和页面元素的操作
    测试登录界面
    在测试框架中使用Log4J 2
    泡泡一分钟:Geometric and Physical Constraints for Drone-Based Head Plane Crowd Density Estimation
    Look Further to Recognize Better: Learning Shared Topics and Category-Specific Dictionaries for Open-Ended 3D Object Recognition
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5217571.html
Copyright © 2011-2022 走看看