HDU 1402 A * B Problem Plus (FFT)

HDU 1402 A * B Problem Plus (FFT)

Question: Calculate the large number A * B.

Idea: Because A and B are too large, java or large numbers will time out. We can use the so-called Fast Fourier Transform (FFT) to solve the polynomial multiplication problem.

For details, see the code:

 

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               #define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const ld eps = 1e-9;const double PI = acos(-1.0);const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;// & 0x7FFFFFFFconst int seed = 131;const ll INF64 = ll(1e18);const int maxn = 200010;int T,n,m, sum[maxn];char s1[maxn/2], s2[maxn/2];struct node { double x, y; node(double x = 0, double y = 0) : x(x), y(y) {} node operator - (const node& b) const { return node(x - b.x, y - b.y); } node operator + (const node& b) const { return node(x + b.x, y + b.y); } node operator * (const node& b) const { return node(x * b.x - y * b.y, x * b.y + y * b.x); }}x1[maxn], x2[maxn];void change(node y[], int len) { int i, j, k; for(i = 1, j = len/2; i < len-1; i++) { if(i < j) swap(y[i], y[j]); k = len/2; while(j >= k) { j -= k; k /= 2; } if(j < k) j += k; }}void fft(node y[], int len, int on) { change(y, len); for(int h = 2; h <= len; h <<= 1) { node wn(cos(-on*2*PI/h), sin(-on*2*PI/h)); for(int j = 0; j < len; j+=h) { node w(1, 0); for(int k = j; k < j+h/2; k++) { node u = y[k]; node t = w * y[k+h/2]; y[k] = u + t; y[k+h/2] = u - t; w = w * wn; } } } if(on == -1) for(int i = 0; i < len; i++) y[i].x /= len;}int main() { while(~scanf("%s%s",s1,s2)) { int len1 = strlen(s1); int len2 = strlen(s2); int len = 1; while(len < len1 * 2 || len < len2 * 2) len <<= 1; for(int i = 0; i < len1; i++) x1[i] = node(s1[len1-1-i]-'0', 0); for(int i = len1; i < len; i++) x1[i] = node(0, 0); for(int i = 0; i < len2; i++) x2[i] = node(s2[len2-1-i]-'0', 0); for(int i = len2; i < len; i++) x2[i] = node(0, 0); fft(x1, len, 1); fft(x2, len, 1); for(int i=0;i
               
                 0) --len; for(int i = len; i >= 0; --i) printf("%c",(char)sum[i] + '0'); printf("\n"); } return 0;}