AtCoder Grand Contest 022

A - Diverse Word

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.

A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, atcoder, zscoderand agc are diverse words while gotou and connect aren't diverse words.

Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.

Let X=x1x2…xn and Y=y1y2…ym be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or xj>yj where j is the smallest integer such that xj≠yj.

Constraints

• 1≤|S|≤26
• S is a diverse word.

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Input

Input is given from Standard Input in the following format:

S

Output

Print the next word that appears after S in the dictionary, or -1 if it doesn't exist.

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Sample Input 1

Copy

atcoder

Sample Output 1

Copy

atcoderb

atcoderb is the lexicographically smallest diverse word that is lexicographically larger than atcoder. Note that atcoderb is lexicographically smaller than b.

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Sample Input 2

Copy

abc

Sample Output 2

Copy

abcd

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Sample Input 3

Copy

zyxwvutsrqponmlkjihgfedcba

Sample Output 3

Copy

-1

This is the lexicographically largest diverse word, so the answer is -1.

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Sample Input 4

Copy

abcdefghijklmnopqrstuvwzyx

Sample Output 4

Copy

abcdefghijklmnopqrstuvx

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这个比赛也挺棒的啊，但是A题我就惨了啊

一个字符串存不存在下一个全排列，存在的话输出-1，否则输出他最小的，这个方法很棒，要不然得分三种情况

#include <bits/stdc++.h>
using namespace std;
int a[];
char s[];
int main()
{
    cin>>s;
    int l=;
    for(int i=; s[i]; i++)a[s[i]]=,l++;
    for(int c='z'; c>='a'; c--)
        if(!a[c])s[l++]=c;
    if(next_permutation(s,s+l))
        cout<<s;
    else
        cout<<-;
    return ;
}

B - GCD Sequence

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers.

She thinks that a set S={a1,a2,…,aN} of distinct positive integers is called special if for all 1≤i≤N, the gcd (greatest common divisor) of ai and the sum of the remaining elements of S is not 1.

Nagase wants to find a special set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a special set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.

Constraints

• 3≤N≤20000

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Input

Input is given from Standard Input in the following format:

N

Output

Output N space-separated integers, denoting the elements of the set S. S must satisfy the following conditions :

• The elements must be distinct positive integers not exceeding 30000.
• The gcd of all elements of S is 1, i.e. there does not exist an integer d>1 that divides all elements of S.
• S is a special set.

If there are multiple solutions, you may output any of them. The elements of S may be printed in any order. It is guaranteed that at least one solution exist under the given contraints.

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Sample Input 1

Copy

3

Sample Output 1

Copy

2 5 63

{2,5,63} is special because gcd(2,5+63)=2,gcd(5,2+63)=5,gcd(63,2+5)=7. Also, gcd(2,5,63)=1. Thus, this set satisfies all the criteria.

Note that {2,4,6} is not a valid solution because gcd(2,4,6)=2>1.

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Sample Input 2

Copy

4

Sample Output 2

Copy

2 5 20 63

{2,5,20,63} is special because gcd(2,5+20+63)=2,gcd(5,2+20+63)=5,gcd(20,2+5+63)=10,gcd(63,2+5+20)=9. Also, gcd(2,5,20,63)=1. Thus, this set satisfies all the criteria.

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让你构造一个序列，首先这个序列的值的gcd值为1，其他的每一个数和其他数的和的gcd不是1

这个构造好难啊，看到别人的那个序列才知道还有这种操作

#include<bits/stdc++.h>
using namespace std;
int n,t[];
int main()
{
    scanf("%d",&n);
    if(n==)printf("2 5 63");
    else
    {
        if(n&)
            t[]=,t[]=,t[]=,t[]=,t[]=,t[]=,t[]=,t[]=;
        else
            t[]=,t[]=,t[]=,t[]=,t[]=,t[]=,t[]=,t[]=;
        for(int i=; i<min(n,); i++)printf("%d ",t[i]);
        for(int i=; i<n; i++)t[i%]=t[i%]+,printf("%d ",t[i%]);
    }
    return ;
}