FFT版题 [51 Nod 1028] 大数乘法

题目链接：51 Nod 传送门
数的长度为$10^5$,乘起来后最大长度为$2 imes10^5$
由于FFT需要把长度开到$2$的次幂，所以不能只开到$2 imes10^5$，会TLE（卡了好久，还以为是要压位）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 400005;//!!!
const double Pi = acos(-1.0);

struct complex
{
	double r, i;
	complex(double _r=0, double _i=0):r(_r), i(_i){}
	complex operator +(const complex &t)const
	{
		return complex(r + t.r, i + t.i);
	}
	complex operator -(const complex &t)const
	{
		return complex(r - t.r, i - t.i);
	}
	complex operator *(const complex &t)const
	{
		return complex(r*t.r - i*t.i, r*t.i + t.r*i);
	}
}a1[MAXN], a2[MAXN], w, wn;
char s1[MAXN], s2[MAXN];
int n, len1, len2, ans[MAXN];

inline void change(complex arr[], int len)
{
	for(int i = 1, j = len/2, k; i < len-1; ++i)
	{
		if(i < j) swap(arr[i], arr[j]);
		for(k = len/2; k <= j; j-=k, k>>=1);
		j += k;
	}
}

inline void fft(complex arr[], int len, int flg)
{
	for(int i = 2; i <= len; i<<=1)
	{
		wn = complex(cos(Pi*flg*2/i), sin(Pi*flg*2/i));
		for(int j = 0; j < len; j+=i)
		{
			w = complex(1, 0);
			for(int k = j; k < j + i/2; ++k)
			{
				complex u = w * arr[k + i/2];
				complex v = arr[k];
				arr[k] = v + u;
				arr[k + i/2] = v - u;
				w = w * wn;
			}
		}
	}
	if(flg == -1)
		for(int i = 0; i < len; ++i)
			arr[i].r /= len;
}

inline void FFT(complex arr[], int len, int flg)
{
	change(arr, len);
	fft(arr, len, flg);
}

int main()
{
	scanf("%s", s1), len1 = strlen(s1);
	scanf("%s", s2), len2 = strlen(s2);
	int len = len1 + len2;
	for(n = 1; n < len; n<<=1);

	for(int i = 0; i < len1; ++i) a1[i] = complex((double)(s1[i] - '0'), 0);
	for(int i = len1; i < n; ++i) a1[i] = complex();
	FFT(a1, n, 1);

	for(int i = 0; i < len2; ++i) a2[i] = complex((double)(s2[i] - '0'), 0);
	for(int i = len2; i < n; ++i) a2[i] = complex();
	FFT(a2, n, 1);
	
	for(int i = 0; i < n; ++i) a2[i] = a1[i] * a2[i];
	FFT(a2, n, -1);
	
	for(int i = 0; i < len1+len2-1; ++i)
		ans[i] = (int)(a2[i].r + 0.5);
	
	for(int i = len1+len2-2; i; --i)
	{
		ans[i-1] += ans[i]/10;
		ans[i] %= 10;
	}
	int i;
	for(i = 0; !ans[i] && i < len1+len2-1; ++i);
	if(i == len1+len2-1) putchar('0');
	else while(i < len1+len2-1) printf("%d",ans[i++]); //此处不能用putchar
	//因为我ans[0]没有向前进位，所以要用%d输出
	putchar(10);
}