思路：

（1）题目抽象为求（ns+n*(n+1)/2*d + sum)%m的最小值

（2)由裴属定理，ns+n*(n+1)/2*d = k1*g1(n,n*(n+1)/2);

 (3)所以为求（k1g1 + sum)%m = ans的最小值；

（4）即k1g1 +k2m = ans - sum;

（5）又k1g1 + k2m = k3g2(g1,m);

(6)即求k3g2 + sum = ans,中ans最小值，即sum%g2的最小值，直接取正模即可，这样我们就拿到了ans，再由拓展欧几里得求出k1，再拓展一次求出s,d即可，这时候值可能为负，由于全局对m取模，所以对s,d取正模即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

typedef long long LL;

LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (!b)
    {
        x = 1,y = 0;
        return a;
    }
    LL d = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
    return d;
}

LL gcd(LL a,LL b)
{
    return b? gcd(b,a%b) : a;
}
LL MOD(LL a,LL b)
{
    return (a%b + b)%b;
}

int main()
{
    LL n, m;
    cin >> n >> m;
    LL sum = 0;
    for (int i = 0; i < n; i++)
    {
        LL x;
        cin >> x;
        sum += x;
    }
	
	LL ans,k1,k2,S,D;
	LL g1 = gcd(n,n*(n + 1)/2);
	LL g2 = gcd(g1,m);
	ans = MOD(sum,g2);
	
	g2 = exgcd(g1,m,k1,k2);
	k1 *= ( (ans - sum)/g2 ) % m;
	k1 %= m;
	g1 = exgcd(n,n*(n + 1)/2,S,D);
	S*= k1,D*= k1;
    
    S = MOD(S,m),D = MOD(D,m);
    cout << ans << endl;
    cout << S <<" "<<D;
    
    return 0;
}