It is a low problem

约数之和题解：

#include<bits/stdc++.h> typedef long long ll; const int INF=0x3f3f3f3f; const int MOD=9901; using namespace std; int poww(int a, int b) { int sum = 1; a %= MOD; while(b) { if(b&1) sum = asum%MOD; a = a%MOD * a%MOD; b >>= 1; } return sum%MOD; }

long long sum(int p, int c) { if(c == 0) return 1; if(c&1) return ((1 + poww(p, (c+1)>>1)) sum(p, (c - 1) >> 1))%MOD; else return ((1 + poww(p, c>>1)) sum(p, (c - 2) >> 1) + poww(p, c))%MOD; }

int main() {

ios::sync_with_stdio(false);

cin.tie(0); int a, b, ans = 1;

cin >> a >> b;

int s = 0; for(int i = 2; i <= a; i++) { s = 0; while(a % i == 0)

  	{

s++; a /= i; } if(s) ans = ans sum(i, sb)%MOD; } if(a == 0) cout << 0 <<endl; else cout << ans <<endl; return 0; }

没必要这么麻烦，看我的题解

我也有
#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,k,temp,cnt=0;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
	k=i;
	while(k)
	{
		temp=k%10;
		if(temp==1) 
		cnt++;      
		k/=10;  
	}
	}
	cout<<cnt;
	return 0;
}