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  • [hdu1402]大数乘法(FFT模板)

    题意:大数乘法

    思路:FFT模板

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    /* ******************************************************************************** */
    #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<intint> 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;
    }
    /* ******************************************************************************** */
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4693001.html
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