题意:大数乘法
思路:FFT模板
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 | /* ******************************************************************************** */ #include <iostream> // #include <cstdio> // #include <cmath> // #include <cstdlib> // #include <cstring> // #include <vector> // #include <ctime> // #include <deque> // #include <queue> // #include <algorithm> // #include <map> // #include <cmath> // using namespace std; // // #define pb push_back // #define mp make_pair // #define X first // #define Y second // #define all(a) (a).begin(), (a).end() // #define foreach(a, i) for (typeof(a.begin()) i = a.begin(); i != a.end(); ++ i) // #define foreach(a, n, i) for(typeof(*a) *i = a; i < a + n; i ++) // #define fillchar(a, x) memset(a, x, sizeof(a)) // // void RI(vector< int >&a, int n){a.resize(n); for ( int i=0;i<n;i++) scanf ( "%d" ,&a[i]);} // void RI(){} void RI( int &X){ scanf ( "%d" ,&X);} template < typename ...R> // void RI( int &f,R&...r){RI(f);RI(r...);} void RI( int *p, int *q){ int d=p<q?1:-1; // while (p!=q){ scanf ( "%d" ,p);p+=d;}} void print(){cout<<endl;} template < typename T> // void print( const T t){cout<<t<<endl;} template < typename F, typename ...R> // void print( const F f, const R...r){cout<<f<< ", " ;print(r...);} template < typename T> // void print(T*p, T*q){ int d=p<q?1:-1; while (p!=q){cout<<*p<< ", " ;p+=d;}cout<<endl;} // // typedef pair< int , int > pii; // typedef long long ll; // typedef unsigned long long ull; // // template < typename T> bool umax(T&a, const T&b){ return b>a? false :(a=b, true );} // template < typename T> bool umin(T&a, const T&b){ return b<a? false :(a=b, true );} // template < typename T> // void V2A(T a[], const vector<T>&b){ for ( int i=0;i<b.size();i++)a[i]=b[i];} // template < typename T> // void A2V(vector<T>&a, const T b[]){ for ( int i=0;i<a.size();i++)a[i]=b[i];} // // const double PI = acos (-1); // // /* -------------------------------------------------------------------------------- */ namespace FFT { const static int maxn = 5e4 + 7; #define L(x) (1 << (x)) double ax[maxn << 2], ay[maxn << 2], bx[maxn << 2], by[maxn << 2]; //需要四倍空间 int revv( int x, int bits) { int ret = 0; for ( int i = 0; i < bits; i++) { ret <<= 1; ret |= x & 1; x >>= 1; } return ret; } void fft( double * a, double * b, int n, bool rev) { int bits = 0; while (1 << bits < n) ++bits; for ( int i = 0; i < n; i++) { int j = revv(i, bits); if (i < j) swap(a[i], a[j]), swap(b[i], b[j]); } for ( int len = 2; len <= n; len <<= 1) { int half = len >> 1; double wmx = cos (2 * PI / len), wmy = sin (2 * PI / len); if (rev) wmy = -wmy; for ( int i = 0; i < n; i += len) { double wx = 1, wy = 0; for ( int j = 0; j < half; j++) { double cx = a[i + j], cy = b[i + j]; double dx = a[i + j + half], dy = b[i + j + half]; double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx; a[i + j] = cx + ex, b[i + j] = cy + ey; a[i + j + half] = cx - ex, b[i + j + half] = cy - ey; double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx; wx = wnx, wy = wny; } } } if (rev) { for ( int i = 0; i < n; i++) a[i] /= n, b[i] /= n; } } int solve( int a[], int na, int b[], int nb, int ans[]) { int len = max(na, nb), ln; for (ln = 0; L(ln) < len; ++ln); len = L(++ln); for ( int i = 0; i < len ; ++i) { if (i >= na) ax[i] = 0, ay[i] = 0; else ax[i] = a[i], ay[i] = 0; } fft(ax, ay, len, 0); for ( int i = 0; i < len; ++i) { if (i >= nb) bx[i] = 0, by[i] = 0; else bx[i] = b[i], by[i] = 0; } fft(bx, by, len, 0); for ( int i = 0; i < len; ++i) { double cx = ax[i] * bx[i] - ay[i] * by[i]; double cy = ax[i] * by[i] + ay[i] * bx[i]; ax[i] = cx, ay[i] = cy; } fft(ax, ay, len, 1); for ( int i = 0; i < len; ++i) ans[i] = ( int )(ax[i] + 0.5); return len; } #undef L(x) } const int maxn = 5e4 + 7; char s1[maxn], s2[maxn]; int x[maxn], y[maxn], ans[maxn << 2]; int main() { #ifndef ONLINE_JUDGE freopen ( "in.txt" , "r" , stdin); #endif // ONLINE_JUDGE while (~ scanf ( "%s" , s1)) { scanf ( "%s" , s2); int len1 = strlen (s1), len2 = strlen (s2); for ( int i = 0; i < len1; i ++) x[i] = s1[len1 - i - 1] - '0' ; for ( int i = 0; i < len2; i ++) y[i] = s2[len2 - i - 1] - '0' ; fillchar(ans, 0); int len = FFT::solve(x, len1, y, len2, ans), i; for (i = 0; i < len || ans[i] >= 10; i ++) { ans[i + 1] += ans[i] / 10; ans[i] %= 10; } len = i; while (ans[len] <= 0 && len) len --; for ( int i = len; i >= 0; i --) putchar (ans[i] + '0' ); puts ( "" ); } return 0; } /* ******************************************************************************** */ |