51nod 1138 consecutive integers of the and (mathematics)

Title Description:

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1138

A positive integer n is given, and N is written as a number of consecutive numbers and forms (length >= 2). For example, n = 15, can be written as 1 + 2 + 3 + 4 + 5, can also be written as 4 + 5 + 6, or 7 + 8. If it cannot be written as a number of consecutive integers, the output is no solution.

Input

Enter 1 number n (3 <= n <= 10^9).

OutPut

Outputs the 1th number in a continuous integer, and if there are more than one order of ascending order, the output no solution if it cannot be decomposed into a number of consecutive integers.

Input Example

15

Output Example

147 Topic Analysis: This problem has a lot of math skills, I was in the "Short Code of the United States" to learn the solution, please see the "Short Code of the United States." AC Code:
/**  * @xiaoran * *  #include <iostream> #include <cstdio> #include <map> #include <cstring > #include <string> #include <algorithm> #include <queue> #include <vector> #include < stack> #include <cstdlib> #include <cctype> #include <cmath> #define LL long longusing namespace std; int main () {int N;while (cin>>n) {        stack<int> st;        int i=1;        while (n-i>1) {            n-=i++;            if (n%i==0) {                //cout<<n/i<<endl;                St.push (n/i);            }        }        if (St.empty ()) {            cout<< "No solution" <<endl;            Continue;        }        while (!st.empty ()) {            cout<<st.top () <<endl;            St.pop ();        }} return 0;}

Appended: The number of sequential sequences that comprise N, and the composition of these sequences.

/** * Number of consecutive numbers, short code of the beauty problem * and find these combinations */#include <iostream>using namespace std;void print (int n) {//print these sequence combinations    int i=0,s= n;    while (n-i>0) {        n-=i++;        if (n%i==0) {            cout<<s<< "=";            int k,k1;            for (k=0,k1=n/i;k<i-1;k++,k1++) {                cout<<k1<< "+";            }            cout<<k1<<endl;        }}}    int main () {    int n;    while (cin>>n) {        int sum=0,i=0;        print (n);        while (n>0) {            n-=++i;            n%i| | sum++;        }        cout<<sum<<endl;    }    return 0;}

51nod 1138 consecutive integers of the and (mathematics)