The beauty of programming: finding integers that match the criteria

I. Description of the problem

Arbitrarily given a positive integer n, a minimum positive integer M (m>1) is obtained so that the N*m decimal form contains only 1 and 0.

For example N=99,m=1 122 334 455 667 789, n*m=111 111 111 111 111 111;

M is the number that meets the criteria when n=99

Ii. Thinking of solving problems

Consider turning the problem into: finding the smallest positive integer that contains only 0 and 1 that can be divisible by N. It can be seen that this is equivalent to the original problem.

So is it necessary to traverse all the numbers that comprise 0 and 1 from small to large? In this case, if the number is looking for k, you need to search the number of 2k-1 to get results.

The method used here is to preserve the remainder information of the%n in the calculation and avoid unnecessary computations. More formalized discourse:

If all K-bit (X) decimal numbers have been traversed, and the minimum number of k+1 digits of t=10k (10) is also searched, all k+1 numbers (Y) are now examined. The

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Y=x+t (that is, the number of all K-Systems +10k), if we divide the space by the X by%n, and the X is decomposed into the equivalence class of the remainder (0~n-1), then the search Y is only required to take the representative element in x to modulo, thus reducing the search time from 2K to N. The smallest element is saved in each equivalence class in a specific implementation.

Third, code implementation

#include <iostream> #include <vector> #include <string> #define N 100213 using namespace std;
Vector<vector<int> >BigIntVec;
    void Printnum (const vector<int>& TV) {//cout<< "print" <<endl;
    int Maxindex=tv.back ();
   
    String Numstr= "";
    for (int i=0;i<maxindex+1;i++) {numstr+= "0";
    for (int i=0;i<tv.size (); i++) {numstr[maxindex-tv[i]]= ' 1 ';
    cout<< "The minimum number of eligible numbers found is:" <<endl;
    for (int i=numstr.size () -1;i>=0;i--) {cout<<numstr[i];
} cout<<endl;
        } void FindNum () {for (int i=0;i<n;i++) {vector<int>tt;
    Bigintvec.push_back (TT);
    } bigintvec[1].push_back (0);
    int noupdate=0; for (int i=1,j=10%n;;;
        i++,j= (j*10)%N) {bool Flag=false;
            if (Bigintvec[j].size () ==0) {bigintvec[j].push_back (i);
        Flag=true;
               
 for (int k=0;k<n;k++) {           if (Bigintvec[k].size () >0&&i>bigintvec[k].back ()) {int t= (K+J)
                   
                %N;
                        if (Bigintvec[t].size () ==0) {for (int tt=0;tt<bigintvec[k].size (); tt++) {
                    Bigintvec[t].push_back (Bigintvec[k][tt]);
                    } bigintvec[t].push_back (i);
                Flag=true;
        }} if (Flag==false) {noupdate++;
        }else{noupdate=0; } if (Bigintvec[0].size () >0| |
        Noupdate==n) {break;
    } if (Bigintvec[0].size () >0) {printnum (bigintvec[0]);
    }else{cout<< "does not find a qualifying number" <<endl;
    int main () {findnum ()};
    System ("pause");
return 0; }

Note: Because the integer involved in this issue can be very large and cannot be represented by a built-in type int or long, vector is used to simulate integers in the program. Because the number of searches is only 1, 2 kinds of numbers, in order to save space, each integer is represented by vector<int>, and vector each element holds 1 occurrences. For example, the number 100101, vector<int> is expressed as {0,2,5}, where 1 of the positions are No. 0, 2, 5 digits respectively.

Iv. output of the program (Input N: 100213):