#include <stdio.h>     // this solution is dynamic programming using memory cutting!
	
	int n, i, front, cnt, sum;
	
	main()
	{
		scanf("%d",&n);
	
		for(i=1;i<=n;i++)
		{
			sum+=i;
			while(sum>n)
				sum-= ++front;
			if(sum==n) cnt++;
		}
		printf("%d",cnt);
	}

	

	#include <stdio.h>      // This is methmatical skill
	
	long long N;
	int count;
	
	
	int main(){
	    long long k;
	
	    scanf("%lld",&N);
	    for(k=1;k<=N;k++){
	        if( (2*N-k*k+k)%(2*k) ==0 && (2*N-k*k+k)/(2*k)>=1) count++;
	    }
	    printf("%d",count);
	
	    return 0;
	}
	

	#include<stdio.h>				// This is SCC(Strongly Connected Component). Try seach SCC (using googling!!)

	#include<vector>
	using namespace std;
	
	vector<int> way[10005];
	vector<int> bay[10005];
	vector<int> nay[10005];
	vector<int> stack;
	
	int num[10005],ad[10005],dy[10005],n,m,st,en,cnt=0;
	
	bool check[10005],sch,ech;
	
	void input()
	{
		int i,a,b;
		scanf("%d%d%d%d",&n,&m,&st,&en);
		for(i=0;i<m;i++){
			scanf("%d%d",&a,&b);
			way[a].push_back(b);
			bay[b].push_back(a);
		}
	}
	
	void DFS(int x)
	{
		int i;
		for(i=0;i<way[x].size();i++){
			if(check[way[x][i]])continue;
			check[way[x][i]]=true;
			DFS(way[x][i]);
		}
		stack.push_back(x);
	}
	
	void nDFS(int x)
	{
		if(x==st&&!sch){ st=cnt; sch=true; }
		if(x==en&&!ech){ en=cnt; ech=true; }
		int i;
		for(i=0;i<bay[x].size();i++){
			if(ad[bay[x][i]]!=0&&ad[bay[x][i]]!=cnt&&(nay[ad[bay[x][i]]].empty()||nay[ad[bay[x][i]]].back()!=cnt)){
				nay[ad[bay[x][i]]].push_back(cnt);
			}
			if(check[bay[x][i]])continue;
			check[bay[x][i]]=true;
			nDFS(bay[x][i]);
		}
		num[cnt]++;
		ad[x]=cnt;
	}
	
	void scc()
	{
		int i;
		for(i=1;i<=n;i++){
			if(check[i])continue;
			check[i]=true;
			DFS(i);
		}
		for(i=1;i<=n;i++)check[i]=false;
		for(i=stack.size()-1;i>=0;i--){
			if(check[stack[i]])continue;
			cnt++;
			check[stack[i]]=true;
			nDFS(stack[i]);
		}
	}
	
	void solve()
	{
		int i,j;
		dy[st]=num[st];
		for(i=st;i<=en;i++){
			for(j=0;j<nay[i].size();j++){
				if(dy[nay[i][j]]<num[nay[i][j]]+dy[i])dy[nay[i][j]]=num[nay[i][j]]+dy[i];
			}
		}
		printf("%d",dy[en]);
	}
	
	int main()
	{
		input();
		scc();
		solve();
		return 0;
	}
	

	#include<stdio.h>						// This is Dynamic Programming (named Bitonic Tour) best solution!!
	#define max(a,b) (a>b?(a):(b))
	int d[220][110][110],n,m;
	int s[110][110];
	main()
	{
		int i,j,k;
		scanf("%d%d",&m,&n);
		for(i=1;i<=n;i++) for(j=1;j<=m;j++) scanf("%1d",&s[i][j]);
		for(i=1;i<=n;i++) for(j=1;j<=m;j++)	s[i][j]=s[i][j]==1?1:s[i][j]==0?0:-1;
		for(k=2,d[1][1][1]=s[1][1]+1;k<n+m;k++) for(i=1;i<=n;i++)
		{
			if(s[i][k-i+1]==-1) continue;
			for(j=i;j<=n;j++)
			{
				int score=0;
				if(i!=j) score = s[i][k-i+1]+s[j][k-j+1];
				else score = s[i][k-i+1];
				if(s[j][k-j+1]==-1) continue;
				d[k][i][j] = max(d[k-1][i-1][j],max(d[k-1][i][j-1],max(d[k-1][i][j],d[k-1][i-1][j-1])));
				d[k][i][j] = d[k][i][j]?d[k][i][j]+score:0;
			}
		}
		printf("%d\n",d[n+m-1][n][n]?d[n+m-1][n][n]-1:0);
	}