考虑只有一个损失时，损失值的生成函数为$A(x)$。

如果不考虑无序方案，有两个损失的生成函数为$B(x)=A(x)A(x)$，同理有三个的时候$C(x)=A(x)A(x)A(x)$。

考虑如何得到无序方案：

选择两个的时候：

$ab$的排列有$ab,ba$两种，我们先减去$aa,bb$的然后除以二就是$B(x)$了，所以$B(x)=\frac{A(x)A(x)-D(x)}{2}$。

选择三个的时候：

$abc$的排列共有$6$种，同样的我们先减去$aaa,bbb,ccc$这样相同的，然后除以$6$就是$C(x)=\frac{A(x)A(x)A(x)-E(x)}{2}$。

#include <bits/stdc++.h>using namespace std;struct Complex {double r, i;Complex(double _r = 0, double _i = 0) : r(_r), i(_i) {}
};Complex operator + (const Complex &a, const Complex &b) {return Complex(a.r + b.r, a.i + b.i);
}Complex operator - (const Complex &a, const Complex &b) {return Complex(a.r - b.r, a.i - b.i);
}Complex operator * (const Complex &a, const Complex &b) {return Complex(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r);
}Complex operator / (const Complex &a, const Complex &b) {return Complex((a.r * b.r + a.i * b.i) / (b.r * b.r + b.i * b.i), (a.i * b.r - a.r * b.i) / (b.r * b.r + b.i * b.i));
}Complex operator * (const Complex &a, const double &b) {return Complex(a.r * b, a.i * b);
}typedef long long ll;const int N = 3e5 + 10;int r[N];Complex x[N], y[N], z[N], ans[N];void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}void FFT(Complex *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}const double pi = acos(-1.0);for (int mid = 1; mid < lim; mid <<= 1) {Complex wn = Complex(cos(pi / mid), rev * sin(pi / mid));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {Complex w = Complex(1, 0);for (int k = 0; k < mid; k++, w = w * wn) {Complex x = f[cur + k], y = w * f[cur + mid + k];f[cur + k] = x + y, f[cur + mid + k] = x - y;}}}if (rev == -1) {for (int i = 0; i < lim; i++) {f[i].r /= lim;}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n;scanf("%d", &n);for (int i = 1, a; i <= n; i++) {scanf("%d", &a);x[a].r++, y[a + a].r++, z[a + a + a].r++;} int lim = 1;while (lim <= 3 * 40000) {lim <<= 1;}get_r(lim);FFT(x, lim, 1), FFT(y, lim, 1), FFT(z, lim, 1);for (int i = 0; i < lim; i++) {ans[i] = ans[i] + (x[i] * x[i] * x[i] - 3.0 * x[i] * y[i] + 2 * z[i]) * (1.0 / 6.0);ans[i] = ans[i] + (x[i] * x[i] - y[i]) * (1.0 / 2);ans[i] = ans[i] + x[i];}FFT(ans, lim, -1);for (int i = 0; i < lim; i++) {int res = int(ans[i].r + 0.5);if (res) {printf("%d %d\n", i, res);}}return 0;
}