51nod 1138 连续整数的和

#include <bits/stdc++.h>using namespace std;int ans[100100];int main(){int n,limit,i,lst,anssum;while(cin>>n){limit=sqrt(2*n);anssum=0;for(i=2;i<=limit;i++){lst=(0+i-1)*i/2;if((n-lst)%i==0){ans[anssum++]=(n-lst)/i;}}if(!anssum)printf("No Solution\n");else{for(i=anssum-1;i>=0;i--)printf("%d\n",ans[i]);}}}