COCI2016/2017 Round2T4 Prosjecni

題目

分析

這種構造題除了流毒我也不想說什麼。
題解也是直接給了one of the possible solutions，我能做的就是證明這種solution是對的，，，

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首先，只有$n=2$時無解，否則可以構造以下方陣（$n\geq 1$）：
$\begin{matrix} 1 & 2 & \cdots & n-1 & \dfrac{1}{2}n(n-1)\\\\ 1+\dfrac{n(n-1)}{2} & 2+\dfrac{n(n-1)}{2} & \cdots & n-1+\dfrac{n(n-1)}{2} & n(n-1)\\\\ 1+n(n-1) & 2+n(n-1) & \cdots & n-1+n(n-1) & \dfrac{3}{2}n(n-1)\\\\ \vdots & \vdots & \ddots & \vdots & \vdots\\\\ 1+\dfrac{n-2}{2}n(n-1) & 2+\dfrac{n-2}{2}n(n-1) & \cdots & n-1+\dfrac{n-2}{2}n(n-1) & \dfrac{n-1}{2}n(n-1)\\\\ n\left[1+\dfrac{n-2}{2}n(n-1)\right]-\sum\limits_{i=1}^{n-1}\left[1+\dfrac{i-1}{2}n(n-1)\right] & n\left[2+\dfrac{n-2}{2}n(n-1)\right]- \sum\limits_{i=1}^{n-1}\left[2+\dfrac{i-1}{2}n(n-1)\right] & \cdots & n\left[n-1+\dfrac{n-2}{2}n(n-1)\right]- \sum\limits_{i=1}^{n-1}\left[n-1+\dfrac{i-1}{2}n(n-1)\right] & n\left[\dfrac{n-1}{2}n(n-1)\right]- \sum\limits_{i=1}^{n-1}\left[\dfrac{i}{2}n(n-1)\right] \end{matrix}$

文字敘述我不想打了，英文很簡單可以自己看：

• In the first row, we write down the numbers $1,2,...,n - 1,\dfrac{n(n-1)}{2}$.
• Each following row but the last is obtained by adding $\dfrac{n(n-1)}{2}$ to each number from
  the previous row.
• The last row is obtained in the following way:
  For each column, if the numbers $a_1,...,a_{n-1}$ are the numbers written in that column so far, we write the number $na_{n-1}-(a_1 + a_2+ ... + a_{n-1})$ in the nth row in that column. By doing this, we achieved that the average of that column is equal to the next to last number in the column.

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計算可得除了最後一行，每行的平均數是最後一個。
計算可得最後一行的平均數是這行的倒數第二個。
計算可得每列的平均數是這列的倒數第二個。

代碼

#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;

int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+(c^48),c=getchar();
    return x*f;
}

#define MAXN 100
int N;
int Ans[MAXN+5][MAXN+5];

int main(){
    freopen("prosjecni.in" ,"r", stdin);
    freopen("prosjecni.out","w",stdout);
    N=read();
    if(N==2)
        puts("-1");
    for(int i=1;i<=N-1;i++)
        Ans[1][i]=i;
    Ans[1][N]=N*(N-1)/2;
    for(int i=2;i<=N-1;i++)
        for(int j=1;j<=N;j++)
            Ans[i][j]=Ans[i-1][j]+N*(N-1)/2;
    for(int i=1;i<=N;i++){
        int Sum=0;
        for(int j=1;j<=N-1;j++)
            Sum+=Ans[j][i];
        Ans[N][i]=N*Ans[N-1][i]-Sum;
    }
    for(int i=1;i<=N;i++,puts(""))
        for(int j=1;j<=N;j++)
            printf("%d ",Ans[i][j]);
}