题解 - Vijos

8

#py大法好
a = int(input())
b = int(input())
print(a*b)

JimmyChen @ 7 年前

无敌
· @ 7 年前

666
海之精华 @ 6 年前

66666666
yangjz1125 @ 6 年前

python 同道中人
蒟蒻在此-QwQ- @ 5 年前

666
蒟蒻在此-QwQ- @ 5 年前

orz
NickWllde @ 5 年前
print(int(input()) * int(input()) )
Copy
明明这样就好了，还要写三行？

6

Sky1231 (sky1231) LV 10 @ 7 年前

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <vector>
#include <deque>
#include <set>
#include <limits>
#include <string>
#include <sstream>
using namespace std;

const int oo_min=0xcfcfcfcf,oo_max=0x3f3f3f3f;

struct bigint
{
    vector<int> num;
    int operator == (bigint b)
    {
        if (num.size()!=b.num.size())
            return 0;
        else
        {
            for (unsigned long long i=0,size=num.size();i<size;i++)
                if (num[i]!=b.num[i])
                    return 0;
            return 1;
        }
    }
    int operator != (bigint b)
    {
        return ((!(*this==b))?1:0);
    }
    int operator < (bigint b)
    {
        if (num.size()<b.num.size())
            return 1;
        else if (num.size()>b.num.size())
            return 0;
        else
        {
            for (unsigned long long i=0,size=num.size();i<size;i++)
            {
                if (num[size-1-i]<b.num[size-1-i])
                    return 1;
                else if (num[size-1-i]>b.num[size-1-i])
                    return 0;
            }
            return 0;
        }
    }
    int operator > (bigint b)
    {
        if (num.size()<b.num.size())
            return 0;
        else if (num.size()>b.num.size())
            return 1;
        else
        {
            for (unsigned long long i=0,size=num.size();i<size;i++)
            {
                if (num[size-1-i]<b.num[size-1-i])
                    return 0;
                else if (num[size-1-i]>b.num[size-1-i])
                    return 1;
            }
            return 0;
        }
    }
    int operator <= (bigint b)
    {
        return ((!(*this>b))?1:0);
    }
    int operator >= (bigint b)
    {
        return ((!(*this<b))?1:0);
    }
    bigint mov(long long x)
    {
        bigint ans;
        ans.num.clear();
        if (x+num.size()>0)
        {
            ans.num.resize(num.size()+x,0);
            for (long long i=0,size=num.size();i<size;i++)
                if (i+x>=0)
                    ans.num[i+x]=num[i];
        }
        else
            ans=0;
        return ans;
    }
    bigint operator = (long long b)
    {
        char s[20+1];
        sprintf(s,"%lld",b);
        num.clear();
        num.resize(strlen(s));
        for (unsigned long long i=0,size=num.size();i<size;i++)
            num[i]=(s[size-1-i]-'0');
        return (*this);
    }
    bigint operator = (string b)
    {
        num.clear();
        num.resize(b.size());
        for (unsigned long long i=0,size=num.size();i<size;i++)
            num[i]=(b[size-1-i]-'0');
        return (*this);
    }
    bigint operator + (bigint b)
    {
        bigint ans;
        ans.num.clear();
        ans.num.resize(max(num.size(),b.num.size()));
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            ans.num[i]=((i<num.size())?num[i]:0)+((i<b.num.size())?b.num[i]:0);
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
        {
            if (ans.num[i]>=10&&i==size-1)
                ans.num.resize(++size);
            ans.num[i+1]+=(ans.num[i]/10);
            ans.num[i]%=10;
        }
        return ans;
    }
    bigint operator += (bigint b)
    {
        return (*this=(*this+b));
    }
    bigint operator - (bigint b)
    {
        bigint ans;
        if ((*this)==b)
        {
            ans=0;
            return ans;
        }
        ans.num.clear();
        ans.num.resize(max(num.size(),b.num.size()));
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            ans.num[i]=((i<num.size())?num[i]:0)-((i<b.num.size())?b.num[i]:0);
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            while (ans.num[i]<0)
                ans.num[i]+=10,ans.num[i+1]--;
        while (ans.num[ans.num.size()-1]==0)
            ans.num.resize(ans.num.size()-1);
        return ans;
    }
    bigint operator -= (bigint b)
    {
        return (*this=(*this-b));
    }
    bigint operator * (bigint b)
    {
        bigint ans,num_0;
        num_0=0;
        ans.num.clear();
        if (*this!=num_0&&b!=num_0)
        {
            ans.num.resize(num.size()+b.num.size()-1);
            for (unsigned long long i=0,size_1=num.size();i<size_1;i++)
                for (unsigned long long j=0,size_2=b.num.size();j<size_2;j++)
                    ans.num[i+j]+=(num[i]*b.num[j]);
            for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            {
                if (ans.num[i]>=10&&i==size-1)
                    ans.num.resize(++size);
                ans.num[i+1]+=(ans.num[i]/10);
                ans.num[i]%=10;
            }
        }
        else
            ans=0;
        return ans;
    }
    bigint operator *= (bigint b)
    {
        return (*this=(*this*b));
    }
    bigint operator / (bigint b)
    {
        bigint ans,num_0;
        num_0=0;
        if (*this<b)
            return num_0;
        if (*this!=num_0&&*this>num_0)
        {
            bigint x,y;
            x=*this,y=b.mov(num.size()-b.num.size());
            ans.num.resize(num.size()-b.num.size()+1,0);
            for (long long i=ans.num.size()-1;i>=0;i--,y=y.mov(-1))
                while (x>=y)
                    x-=y,ans.num[i]++;
            for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            {
                if (ans.num[i]>=10&&i==size-1)
                    ans.num.resize(++size);
                ans.num[i+1]+=(ans.num[i]/10);
                ans.num[i]%=10;
            }
            while (ans.num.size()>1&&ans.num[ans.num.size()-1]==0)
                ans.num.resize(ans.num.size()-1);
        }
        else
            ans=0;
        return ans;
    }
    bigint operator /= (bigint b)
    {
        return (*this=(*this/b));
    }
    void print()
    {
        for (unsigned long long i=0,size=num.size();i<size;i++)
            printf("%d",num[size-1-i]);
    }
};

char s[10000+1];
#define s(x) s[x-1]
bigint a,b,c;
string s_a,s_b;
#define s_a(x) s_a[x-1]
#define s_b(x) s_b[x-1]

int main()
{
    while (~scanf("%s",s))
    {
        s_a=s;
        getchar();
        scanf("%s",s);
        s_b=s;
        getchar();
        a=s_a,b=s_b;
        c=a*b;
        c.print();
        printf("\n");
    }
}

sky1231 @ 7 年前

很H2O的題
Spencer Moore @ 7 年前

DALAO
ttbr5145 @ 6 年前

@sky1231: 大整数类？

1

beishu LV 10 @ 10 个月前

Java和Python都支持处理大数，就是对C++ OIer不太友好

Java AC code

import java.util.*;
import java.math.*;
public class Main
{
    public static void main(String args[])
    {
        Scanner cin=new Scanner(System.in);
        BigInteger a=cin.nextBigInteger();
        BigInteger b=cin.nextBigInteger();
        System.out.print(a.multiply(b));
    }
}

Python AC code

print(int(input())*int(input()))

只有用C++AC的才有实力

0

15817382763 LV 4 @ 1 个月前

至少对于这种很直白的题python还是很爽的。。。。。。

print(int(input())*int(input()))

0

__cby__ LV 7 @ 2 年前

python的巅峰时刻:

n=int(input())
m=int(input())
print(x*y)

0

永远的天狼星 LV 6 @ 2 年前

a = int(input())

b = int(input())

print(a*b)

0

永远的天狼星 LV 6 @ 2 年前

d = int(input())
f = int(input())
print(d*f)
//ha,ha,用c++的手都要残了

0

黄青玲 LV 7 @ 3 年前

py 无敌！自带高精度！

a = int(input())
b = int(input())
print(a*b)

0

keuin LV 8 @ 4 年前

#include<bits/stdc++.h>
#define ll long long int
using namespace std;
struct bigint{
    int a[7000],n,N=1000,t=3;
    bigint (){
        memset(a,0,sizeof(a));
    }
    bigint(char* s,int _n){
        memset(a,0,sizeof(a));
        char* _end=s+_n,*beg;
        int p=0;
        while(_end-t>s){
            beg=_end-t;
            for(int i=0;i<t;i++){
                a[p]*=10;
                a[p]+=beg[i]-'0';
            }
            ++p;
            _end=beg;
        }
        if(_end>s){
            for(int i=0;i<_end-s;i++){
                a[p]*=10;
                a[p]+=s[i]-'0';
            }
            ++p;
        }
        n=p;
    }
    int& operator [](int i){
        return a[i];
    }
    bigint operator * (bigint& b){
        int i,j,m=b.n;
        bigint r;
        int sr=n+m-1;
        for(i=0;i<n;++i)
            for(j=0;j<m;++j)
                r[i+j]+=a[i]*b[j];
        for(i=0;i<sr;++i){
            r[i+1]+=r[i]/N;
            r[i]%=N;
        }
        if(r[sr])sr++;
        r.n=sr;
        return r;
    }
    void show(){
        int i=n-2;
        printf("%d",a[n-1]);
        while(i>=0)
            printf("%03d",a[i--]);
    }
};

char s[10001];
int main() {
    //freopen("in.txt","r",stdin);
   scanf("%s",s);
   bigint a(s,strlen(s));
   scanf("%s",s);
   bigint b(s,strlen(s));
   (a*b).show();
}

后两个wa了的，可能是中间爆int了，压四位先乘了再进位会爆int...要不用longlong 要不就压三位

0

VL739 LV 4 @ 4 年前

#include<iostream>
#include<string>
using namespace std;
const int N = 10001;
void h_multiply_2(int a[], int b[], int c[])
{
int la = a[0];
int lb = b[0];
for(int i=1;i<=la;i++)
{
for(int j=1;j<=lb;j++)
{
c[i+j-1] += a[i] * b[j];
}
}
for(int i=1;i<=la+lb;i++)
{
c[i+1] += c[i] / 10;
c[i] %= 10;
}
c[0] = la+lb;
}
int convert(string s,int n[])
{
int length = s.length();
for(int i=1;i<=length;i++)
{
n[i] = s[length-i] - '0';
}
n[0] = length;
return length;
}
void output(int n[])//n是算好了的高精度数
{
int length = n[0];
if(length==0)
{
cout << 0 << endl;
return;
}
for(int i=length;i>=1;i--)
{
cout << n[i];
}
cout << endl;
}
int trim(int n[])
{
int length = 0;
for(int i=n[0];i>0;i--)
{
if(n[i]!=0)
{
length = i;
break;
}
}
n[0] = length;
return length;
}
int a[N+1];
int b[N+1];
int c2[2*N+1];
int main()
{
string s1, s2;
cin >> s1 >> s2;
convert(s1, a);
convert(s2, b);
h_multiply_2(a,b,c2);
trim(c2);
output(c2);
return 0；
}

0

此人太弱，用户名无法显示 (蒟蒻在此-QwQ-) LV 7 @ 4 年前

x=int(input())
y=int(input())
print(x*y)

真香！！

0

ttbr5145 LV 5 @ 6 年前

告诉你什么叫做流！弊！
Python3三行代码解决：

x=int(input()) #input()用于读入一行字符，int()用于把字串转为数字
y=int(input());#在python中分号可以省略，也可以加上
print(x*y)     #输出a*b，python中有乘方运算符（^）

哇哈哈哈哈哈哈哈哈哈哈哈哈~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//笑出猪叫

来自外太空的路人 @ 4 年前

明明一行就行
甯聿 @ 2 年前

兄弟，Python中的 ^ 符号不是乘方而是异或

0

hack_leo LV 8 @ 6 年前

其实，除了Python,Haskell也可以一行

main = do { a<-readLn; b<-readLn; print (a*b) }

0

lk983773710 LV 7 @ 6 年前

#include <iostream>
#include <cstring>
using namespace std;
int a[10601],b[10601],c[20601];
int main(int argc, char** argv)
{
int la,lb,x=0,o;
string A,B;
cin>>A>>B;
la=A.length();
lb=B.length();
for(int i=1;i<=la;i++)
{
a[i]=A[la-i]-'0';
}
for(int j=1;j<=lb;j++)
{
b[j]=B[lb-j]-'0';
}
for(int k=1;k<=la;k++)
{
for(int l=1;l<=lb;l++)
{
c[k+l-1]=a[k]*b[l]+c[k+l-1]+x;
x=c[k+l-1]/10;
c[k+l-1]=c[k+l-1]%10;
}
c[k+lb]=x;
x=0;
}
o=la+lb;
while(c[o]==0&&o!=0)
{
o--;
}
for(int oo=o;oo>=1;oo--)
{
cout<<c[oo];
}
return 0;
}

0

crucialize LV 4 @ 6 年前

golang大法
```go

packagepa main

import (
"fmt"
"math/big"
)

func main() {

var stra, strb string

fmt.Scanf("%s\n%s", &stra, &strb)

a := big.NewInt(0)
b := big.NewInt(0)
ans := big.NewInt(0)

a.SetString(stra, 10)
b.SetString(strb, 10)

ans.Mul(a, b)

fmt.Println(ans)
}

0

等下。时光请等一下 (天下第一招3) LV 7 @ 6 年前

#include<bits/stdc++.h>
using namespace std;
int main(){
    char st1[10005],st2[10005];
    int a[10005],b[10005],c[20005];
    int i,j,len1,len2;
    cin >> st1 >> st2;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    len1=strlen(st1);
    len2=strlen(st2);
    j=1;
    for (i=len1-1;i>=0;i--)
    {
        a[j++]=st1[i]-'0';
    }
    j=1;
    for (i=len2-1;i>=0;i--)
    {
        b[j++]=st2[i]-'0';
    }

    for (i=1;i<=len1;i++)
        for (j=1;j<=len2;j++)
            c[i+j-1]+=a[i]*b[j];

    for(i=1;i<len1+len2;i++){
        c[i+1]+=c[i]/10;
        c[i]%=10;
    }
    while (!c[i] && i>1){
        i--;
    }
    while (i){
        printf("%d",c[i--]);
    }
    printf("\n");
    return 0;
}

在网上淘得代码，确实是很不错，模拟你现实用笔算的过程，在计算机中模拟出来，真心给个赞。

0

GUARDIANS LV 8 @ 6 年前

#include<bits/stdc++.h>
using namespace std;
int main()
{
char a1[30000],a2[30000];
int a[30000],b[30000],c[30000],la,lb,lc,i,j,x;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
cin>>a1>>a2;
la=strlen(a1);
lb=strlen(a2);
for(i=0;i<=la-1;i++) a[la-i]=a1[i]-48;
for(i=0;i<=lb-1;i++) b[lb-i]=a2[i]-48;
for(i=1;i<=la;i++)
{
x=0;
for(j=1;j<=lb;j++)
{
c[i+j-1]=a[i]*b[j]+x+c[i+j-1];
x=c[i+j-1]/10;
c[i+j-1]%=10;
}
c[i+lb]=x;
}
lc=la+lb;
while(c[lc]==0&&lc>1)
lc--;
for(i=lc;i>=1;i--)
cout<<c[i];
cout<<endl;
}

0

GUARDIANS LV 8 @ 6 年前

#include<bits/stdc++.h>
using namespace std;
int main()
{
char a1[30000],a2[30000];
int a[30000],b[30000],c[30000],la,lb,lc,i,j,x;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(c,0,sizeof(c));
cin>>a1>>a2;
la=strlen(a1);
lb=strlen(a2);
for(i=0;i<=la-1;i++) a[la-i]=a1[i]-48;
for(i=0;i<=lb-1;i++) b[lb-i]=a2[i]-48;
for(i=1;i<=la;i++)
{
x=0;
for(j=1;j<=lb;j++)
{
c[i+j-1]=a[i]*b[j]+x+c[i+j-1];
x=c[i+j-1]/10;
c[i+j-1]%=10;
}
c[i+lb]=x;
}
lc=la+lb;
while(c[lc]==0&&lc>1)
lc--;
for(i=lc;i>=1;i--)
cout<<c[i];
cout<<endl;
}

0

Orina_zju LV 8 @ 7 年前

Java大法好

import java.math.BigInteger;
import java.util.Scanner;

public class Main 
{
    public static void main(String[] args) 
    {
        Scanner in = new Scanner(System.in);
        BigInteger A = in.nextBigInteger(), B = in.nextBigInteger();
        System.out.println(A.multiply(B));
        in.close();
    }
}

0

zcq1412 LV 7 @ 7 年前

print(int(input())int(input()))

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337 条题解

高精度乘法

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337 条题解

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