P1919 【模板】高精度乘法 | A*B Problem 升级版

思路讲解

使用FFT。

注意给FFT喂入的是从低到高的。

输出前注意去除前导0。

string A,B;
cin>>A>>B;
vll a,b;
ROF(i,SZ(A)-1,0){
    a.pb(A[i]-'0');
}
ROF(i,SZ(B)-1,0){
    b.pb(B[i]-'0');
}
vll ans=multiply(a, b);
ll carry=0;
FOR(i,0,SZ(ans)-1){
    ans[i]+=carry;
    carry=ans[i]/10;
    ans[i]%=10;
}
while (!ans.empty() && ans.back()==0) {
    ans.pop_back();
}
if(SZ(ans)==0){
    cout<<0;
    return;
}
ROF(i,SZ(ans)-1,0){
    cout<<ans[i];
}

AC代码

https://www.luogu.com.cn/record/229526218

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) (var<<1|1)
#define fsp(x) fixed<<setprecision(x)
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using cd=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    // while(lT--)
        __();
    return 0;
}
/*

*/
// 理论讲解 https://www.bilibili.com/video/BV1za411F76U/
// 参考（抄的） https://cp-algorithms.com/algebra/fft.html
// 多项式乘法 https://www.luogu.com.cn/record/229525769
using cd=complex<double>;
void fft(vector<cd> &a,bool invert){
    ll n=SZ(a);
    if(n==1) return;
    vector<cd> a0(n/2),a1(n/2); // a0 even a1 odd
    FOR(i,0,n/2-1){
        a0[i]=a[2*i];
        a1[i]=a[2*i+1];
    }
    fft(a0,invert),fft(a1,invert);
    double ang=2*pi/n*(invert?-1:1);
    cd w(1),wlen={cos(ang),sin(ang)};   // wlen 是步进长度
    FOR(i,0,n/2-1){ 
        // A(x)=Ae(x^2)+x*Ao(x^2)   A(-x)=Ae(x^2)-x*Ae(x^2)
        // a[i]=a0[i]+a1[i] -a[i]=a0[i]-a1[i]; 
        // a[i] 和 a[i+n/2] 是单位圆上相反的一对
        a[i]=a0[i]+w*a1[i];
        a[i+n/2]=a0[i]-w*a1[i];
        if(invert){ 
            // 每次都除了2，一共有log2(n)层，所以一共除了2^log2(n)=n
            a[i]/=2;
            a[i+n/2]/=2;
        }
        w*=wlen;
    }
}
vll multiply(vll &a,vll &b){
    vector<cd> fa(all(a)),fb(all(b));
    ll n=1;
    while(n<SZ(a)+SZ(b)){
        n<<=1;
    }
    fa.resize(n);fb.resize(n);
    fft(fa,false);fft(fb,false);
    FOR(i,0,n-1) fa[i]*=fb[i];
    fft(fa,true);
    vll res(n);
    //              避免精度损失
    FOR(i,0,n-1) res[i]=llround(fa[i].real());
    return res;
}
inline void __(){
    string A,B;
    cin>>A>>B;
    vll a,b;
    ROF(i,SZ(A)-1,0){
        a.pb(A[i]-'0');
    }
    ROF(i,SZ(B)-1,0){
        b.pb(B[i]-'0');
    }
    vll ans=multiply(a, b);
    ll carry=0;
    FOR(i,0,SZ(ans)-1){
        ans[i]+=carry;
        carry=ans[i]/10;
        ans[i]%=10;
    }
    while (!ans.empty() && ans.back()==0) {
        ans.pop_back();
    }
    if(SZ(ans)==0){
        cout<<0;
        return;
    }
    ROF(i,SZ(ans)-1,0){
        cout<<ans[i];
    }
}
/*
AC
https://www.luogu.com.cn/record/229526218
*/

心路历程（WA，TLE，MLE……）