Long Long Long int= String ? ? ? ?

Computer Science Level 3

What is the result if you multiply the first 100 positive prime number, input only the first 33 digit of the number :v . Hint : the answer doesnt fit in Qword/Unsigned long long int ...


The answer is 471193079990618495316248783476026.

A Steven Kusuman by A Steven Kusuman , 25, Indonesia

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2 solutions

A Steven Kusuman
Jan 28, 2015 
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#include<stdio.h>

#include<cstring>

#include<iostream>

#include<cmath>

using namespace std;

void itos(int i,string &s){

char c;
if(i>0){
    c=char(i%10+48);
    s=s+c;
    i=trunc(i/10);
    itos(i,s);

}
}
string tam(string s1,string s2)
{

     string has;
     int p1,p2,i;
     int isi,ban=0;

     p1=s1.size(); p2=s2.size();
     has="";
     if(p1<p2) for(i=0;i<(p2-p1);i++) s1="0"+s1; else
     if(p2<p1) for(i=0;i<(p1-p2);i++) s2="0"+s2;
     p1=s1.size();
     for(i=p1-1;i>=0;i--)
     {

        isi=int(s1[i])-48+int(s2[i])-48+ban;
        ban=isi/10;
        isi%=10;
        has=char(isi+48)+has;

     }
     if(ban>0) has=char(ban+48)+has;

     return has;

}

string kal(string s1,string s2)
{

    string has="0",ban="";
    int ang,sim=0;
    int p1=s1.length(),p2=s2.length();

    if (p2 > p1) return kal(s2,s1);
    for (int i=p2-1;i>=0;i--)
    {

        ban="";sim=0;
        for (int j=p1-1;j>=0;j--)
        {

            ang = sim + (int(s2[i])-48)*(int(s1[j])-48);
            sim = ang / 10;
            ang%= 10;
            ban = char(ang+48) + ban;

        }
        if (sim > 0) ban = char(sim+48) + ban;
        for (int j=0;j<(p2-1-i);j++)
            ban+='0';
        //cout<<ban<<endl;
        has=tam(has,ban);

    }

    return has;

}

int main(){

bool arr[1001];
long int arr2[101];
long int x,cnt=0;
memset(arr,true,sizeof(arr));
for (int i=2;i<=trunc(sqrt(1000));i++){

    if(arr[i]==true){

        x=trunc(pow(i,2));
        while(x<=1000){

            arr[x]=false;
            x=x+i;

        }

    }

}

unsigned long long int row=0;
for (long int i=2;i<=1000;i++){

    if(row>99) break;

    if(arr[i]==true){
       arr2[row]=i;
       row++;

    }

}
string xs,xc,cx;
xs=kal("1","1");
for (long int i=0;i<100;i++){

    xc="";
    cx="";
        itos(arr2[i],xc);
for (int ii=xc.length()-1;ii>-1;ii--){

    cx=cx+xc[ii];
}

     xs=kal(xs,cx);
}

cout<<xs;
return 0;

}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Long and not understandable.

Hasmik Garyaka - 3 years, 9 months ago
Brock Brown
Feb 1, 2015

Python: 

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from math import sqrt
def is_prime(x):
    if x < 2:
        return False
    i = 2
    while i <= sqrt(x):
        if x % i == 0:
            return False
        i+=1
    return True
primes = []
guess = 1
product = 1
while len(primes) < 100:
    if is_prime(guess):
        primes.append(guess)
        product *= guess
    guess += 1
print "Answer:", str(product)[0:33]

Evaluation time: 0.01164 seconds

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Python is not challenging enough :)

A Steven Kusuman - 6 years, 4 months ago

Log in to reply

It really isn't, the plain English that it offers makes most algorithms almost too easy to construct and understand.

Brock Brown - 6 years, 4 months ago

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