提交记录 12352


用户 题目 状态 得分 用时 内存 语言 代码长度
axiomofchoice 1004. 【模板题】高精度乘法 Time Limit Exceeded 0 3 s 6 MB + 384 KB C++ 3.14 KB
提交时间 评测时间
2020-03-24 17:59:36 2020-03-24 17:59:43
#include <bits/stdc++.h>
using namespace std;
#define repeat(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define repeat_back(i,a,b) for(int i=(b)-1,_=(a);i>=_;i--)
#define mst(a,x) memset(a,x,sizeof(a))
#define fi first
#define se second
#ifndef qwq
int cansel_sync=[]{ios::sync_with_stdio(0); cin.tie(0); return 0;}();
#endif
const double err=1e-11; typedef long long ll; const int inf=~0u>>2; const ll INF=~0ull>>2; ll read(){ll x; if(scanf("%lld",&x)==-1)exit(0); return x;} typedef double lf; typedef long double llf; const lf pi=acos(-1); lf readf(){lf x; if(scanf("%lf",&x)==-1)exit(0); return x;} template<typename T> T sqr(const T &x){return x*x;} typedef pair<int,int> pii; template<typename A,typename B> ostream &operator<<(ostream &o,const pair<A,B> &x){return o<<'('<<x.fi<<','<<x.se<<')';}
//mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N=200010;
const int mod=(1?1000000007:998244353); ll mul(ll a,ll b,ll m=mod){return a*b%m;} ll qpow(ll a,ll b,ll m=mod){ll ans=1; for(;b;a=mul(a,a,m),b>>=1)if(b&1)ans=mul(ans,a,m); return ans;} ll getinv(ll v,ll m=mod){return qpow(v,m-2,m);}
//#define int ll
struct big{
	vector<ll> a;
	static const int k=100000000;
	int size()const{return a.size();}
	big(){a.push_back(0);}
	explicit big(const ll &x){ //接收ll
		*this=big(to_string(x));
	}
	explicit big(const string &s){ //接收string
		int w=8,len=s.size();
		int f=(s[0]=='-')?-1:1;
		static int p10[]={1,10,100,1000};
		a.resize(len/w+1);
		repeat(i,0,len-(f==-1))
			a[i/w]+=f*(s[len-1-i]-48)*p10[i%w];
		adjust();
	}
	int sgn(){return a.back()>=0?1:-1;} //这个只能在强/弱调整后使用
	void shrink(){ //收缩(内存不收缩)
		while(size()>1 && a.back()==0)a.pop_back();
	}
	void adjust(){ //弱调整
		repeat(i,0,3)a.push_back(0);
		repeat(i,0,size()-1){
			a[i+1]+=a[i]/k;
			a[i]%=k;
		}
		shrink();
	}
	void final_adjust(){ //强调整
		adjust();
		int f=sgn();
		repeat(i,0,size()-1){
			int t=(a[i]+k*f)%k;
			a[i+1]+=(a[i]-t)/k;
			a[i]=t;
		}
		shrink();
	}
	operator string(){ //转换成string
		static char s[10010]; char *p=s;
		final_adjust();
		if(sgn()==-1)*p++='-';
		repeat_back(i,0,size())
			sprintf(p,i==size()-1?"%lld":"%08lld",abs(a[i])),p+=strlen(p);
		return s;
	}
	const ll &operator[](int n)const{ //访问
		return a[n];
	}
	ll &operator[](int n){ //弹性访问
		repeat(i,0,n-size()+1)a.push_back(0);
		return a[n];
	}
};
big operator+(big a,const big &b){
	repeat(i,0,b.size())a[i]+=b[i];
	a.adjust();
	return a;
}
big operator-(big a,const big &b){
	repeat(i,0,b.size())a[i]-=b[i];
	a.adjust();
	return a;
}
big operator*(const big &a,const big &b){
	big ans;
	repeat(i,0,a.size()){
		repeat(j,0,b.size())
			ans[i+j]+=a[i]*b[j];
		ans.adjust();
	}
	return ans;
}
ll operator%(const big &a,ll mod){
	ll ans=0,p=1;
	repeat(i,0,a.size()){
		ans=(ans+p*a[i])%mod;
		p=(p*a.k)%mod;
	}
	return (ans+mod)%mod;
}
bool operator<(big a,big b){
	a.final_adjust();
	b.final_adjust();
	repeat_back(i,0,max(a.size(),b.size()))
		if(a[i]!=b[i])return a[i]<b[i];
	return 0;
}
bool operator==(big a,big b){
	a.final_adjust();
	b.final_adjust();
	repeat_back(i,0,max(a.size(),b.size()))
		if(a[i]!=b[i])return 0;
	return 1;
}
signed main(){
	string a,b;
	cin>>a>>b;
	cout<<string(big(a)*big(b))<<endl;
	return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Testcase #13 s6 MB + 384 KBTime Limit ExceededScore: 0


Judge Duck Online | 评测鸭在线
Server Time: 2020-04-06 10:15:09 | Loaded in 1 ms | Server Status
个人娱乐项目,仅供学习交流使用