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Programming > C++ > Can "cout" do e...
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Can "cout" do evil?

by thomas <FreshThomas@[EMAIL PROTECTED] > May 1, 2008 at 07:48 PM

#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<iterator>
#include<string>
#include<algorithm>

using namespace std;

#define M 1000009

vector<int> primes;

//generate C(n, 1), C(n, 2), ...
void dfs(vector<int> &mp, int its, int i, vector<int> &temp){
    if(i==0) {
        int s=1;
        for(vector<int>::iterator it2(temp.begin()); it2!=temp.end();
it2++){        s=s*(*it2);        }
        if(s>1) mp.push_back(s);   return;
    }
    if(mp[its]==0) return;
    int my_its(its);
    for(vector<int>::iterator it_t(mp.begin() + its); *it_t!=0;  it_t+
+, my_its++){
        temp.push_back(*it_t);
        dfs(mp, my_its+1, i-1, temp);
        temp.pop_back();
    }
}

int main(){
    vector<int> numbers;    numbers.resize(M, 0);

//generate prime numbers
    primes.push_back(2);
    for(int i=3; i<M; i+=2){
        if(numbers[i]==1) continue;
        primes.push_back(i);
        for(int j=i+i; j<M; j+=i)
            numbers[j] = 1;
    }

    int m, k;
    while(cin>>m>>k){
        vector<int> mprime;   int mm=m;
        cout<<1;              //<1>

//prime factors of m
        for(vector<int>::iterator it(primes.begin()); it!=primes.end()
&&*it<=mm; it++){
            if(mm%(*it)==0) mprime.push_back(*it);
            while(mm%(*it)==0) mm=mm/(*it);
        }

//generate C(n,i), push the multiplication result of the "i" factors
in mprime, seperated by 0
        int msize = mprime.size();
        mprime.push_back(0);
        for(int i=2; i<=msize; i++){
            vector<int> temp;
            dfs(mprime, 0, i, temp);
            mprime.push_back(0);
        }

//calculate the k-th number x with gcd(x,m)=1
        int result = 0;   int total=0;  bool change=true;
        while(total<k){
            result += (k-total);    total=result;
            for(vector<int>::iterator it(mprime.begin()); it!
=mprime.end()&&*it<=m&&*it<=result; it++){
                if(*it==0){ change=(change?false:true); continue;}
                if(change)    total-=(result/(*it));
                else total+=(result/(*it));
            }
        }
        cout<<result<<endl;
    }
}

---------code--------
The above is the code to calculate the k-th number x of m having
gcd(x,m)=1
notice the line marked <1>
If the <1> line exists, everything works fine.
But if I remove line <1>, the guy says error.
what's wrong? Can a simple "cout" do anything evil?
 




 5 Posts in Topic:
Can "cout" do evil?
thomas <FreshThomas@[E  2008-05-01 19:48:19 
Re: Can "cout" do evil?
"Alf P. Steinbach&qu  2008-05-02 05:48:27 
Re: Can "cout" do evil?
saya-jin <gyula.gubacs  2008-05-02 03:21:24 
Re: Can "cout" do evil?
"Jim Langston"   2008-05-02 04:42:15 
Re: Can "cout" do evil?
thomas <FreshThomas@[E  2008-05-04 09:22:21 

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tan12V112 Tue Oct 14 11:22:01 CDT 2008.