n children are standing in a circle and playing a game. Children‘s numbers in clockwise order form a permutation a1,?a2,?...,?an of length n. It is an integer sequence such that each integer from 1 to n appears exactly once in it.

The game consists of m steps. On each step the current leader with index i

You are given numbers l1,?l2,?...,?lm — indices of leaders in the beginning of each step. Child with number l1 is the first leader in the game.

Write a program which will restore a possible permutation a

The first line contains two integer numbers n, m (1?≤?n,?m?≤?100).

The second line contains m integer numbers l1,?l2,?...,?lm (1?≤?li?≤?n) — indices of leaders in the beginning of each step.

Print such permutation of n numbers a1,?a2,?...,?an that leaders in the game will be exactly l1,?l2,?...,?lm if all the rules are followed. If there are multiple solutions print any of them.

If there is no permutation which satisfies all described conditions print -1.

4 5
2 3 1 4 4

3 1 2 4 

3 3
3 1 2

-1

Let‘s follow leadership in the first example:

• Child 2 starts.
• Leadership goes from 2 to 2?+?a2?=?3.
• Leadership goes from 3 to 3?+?a3?=?5. As it‘s greater than 4, it‘s going in a circle to 1.
• Leadership goes from 1 to 1?+?a1?=?4.
• Leadership goes from 4 to 4?+?a4?=?8. Thus in circle it still remains at 4.

題意：n個人，進行m次操作，操作規則根據A數組，起始位置為L1 L1+A[L1]得到下一個數字L2

L2+A[L2]得到下一個數字L3這樣進行下去，且Li<=n,如果計算超過則必須%n

現在告訴我們L數組，還原出A數組

解法：

1 不存在情況 A數組數組唯一，若出現重復則不存在

　　　　　　A數字每個位置答案唯一，即如果求出A[2]=3,那麽A[2]不可以等於其他值，若出現其他答案則不存在

2 填補數字 A[i]<A[i+1] 則該位置為A[i+1]-A[i]

　　　　　A[i]>=A[i+1] 則該位置n+A[i+1]-A[i]

未填補數字則缺啥補啥就行，中間註意判斷存在條件

#include<bits/stdc++.h>
using namespace std;
const long long N = 10010;
vector<long long> v;
long long A[N];
long long B[N];
map<int,int>Q;
int main(){
    long long a,b;
    cin >> a >> b;
    for(int i=1;i<=b;i++){
        cin>>A[i];
    }
    for(int i=1;i<=a;i++){
        B[i]=-1;
    }
    long long ans=A[1];
    for(int i=2;i<=b;i++){
        if(A[i]>ans){
            long long temp=A[i]-ans;
            if(B[ans]!=-1&&B[ans]!=temp){
                cout<<"-1";
                return 0;
            }
            B[ans]=temp;
        }else{
            long long num=a+A[i];
            if(B[ans]!=-1&&B[ans]!=num-ans){
                cout<<"-1";
                return 0;
            }
            B[ans]=num-ans;
        }
        ans=A[i];
    }
    for(int i=1;i<=a;i++){
        if(Q[B[i]]>1||B[i]>a){
            cout<<"-1";
            return 0;
        }else if(B[i]!=-1){
            Q[B[i]]++;
        }
    }
    for(int i=1;i<=a;i++){
        if(B[i]==-1){
            long long str=1;
            while(Q[str]) str++;
            Q[str]++;
            B[i]=str;
        }
    }
    for(int i=1;i<=a;i++){
        if(Q[B[i]]>1||B[i]>a){
            cout<<"-1";
            return 0;
        }else if(B[i]!=-1){
            Q[B[i]]++;
        }
    }
    for(int i=1;i<=a;i++){
        cout<<B[i]<<" ";
    }
    return 0;
}

Educational Codeforces Round 24 B