求n = a1*a2*33...*ak 且满足(1<a1<=a2<=a3...<=ak<=n)的状态数

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代码:#include <iostream> using namespace std; int ans; void dfs(int n, int k) { if(n == 1) { ans++; return; } int i; for(i=2; i<=n; ++i) if(n%i==0 && i>=k) dfs(n/i,i); } int main(void) { int n; while(cin >> n) { ans = 0; dfs(n,1); cout << ans << endl; } return 0; }