从拉格朗日插值法还原多项式

从拉格朗日插值法还原多项式_第1张图片

这里有份实现该功能的代码

int Mul(const LL &x,const LL &y) {return x * y % mo;}
int Add(const int &x,const int &y) {return (x + y) % mo;}
int Dec(const int &x,const int &y) {return (x - y + mo) % mo;}

void Work(int now)
{
	int *t1 = B,*t2 = C;
	for (int i = 0; i <= 2 * n; i++) f[i] = Calc(i);
	for (int i = 0; i <= 2 * n; i++)
	{
		for (int j = 0; j <= 2 * n; j++) B[j] = C[j] = 0;
		int tmp = f[i]; t1[0] = 1;
		for (int j = 0; j <= 2 * n; j++)
		{
			if (i == j) continue;
			tmp = Mul(tmp,Inv[i - j]);
			for (int k = 1; k <= 2 * n; k++) t2[k] = t1[k - 1]; t2[0] = 0;
			for (int k = 0; k <= 2 * n; k++) t2[k] = Dec(t2[k],Mul(t1[k],j));
			swap(t1,t2);
		}
		for (int j = 0; j <= 2 * n; j++) D[j] = Add(D[j],Mul(tmp,t1[j]));
	}
}

你可能感兴趣的:(拉格朗日插值法)