HDU6108 小C的倍數問題 【腦洞】

鏈接

---

證明:p進制下,當且僅當,x每位之和爲B的倍數時,x%B=0

設x=k0+k1p+k2p2+...+knpn
若(k0+k1+...+kn)%B=∑ni=0ki%B=0
又因爲x=∑ni=0ki+∑ni=1((pi−1)∗ki)
所以當∑ni=1((pi−1)∗ki)%B=0時,x%B=0
而pi−1=(p−1)∗(pi−1+pi−2+...+1)−>(pi−1)%(p−1)=0   
所以當B=p−1或p−1的因子時,有:當且僅當∑ni=0ki%B=0,x%B=0

#include<stdio.h>
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define MEM(a,x) memset(a,x,sizeof(a))
#define lowbit(x) ((x)&-(x))

using namespace std;

const int N = 5000 + 5;
const double EPS = 1e-7;

int slove(int n){
    int ans=0;
    for(int i=1,ed=sqrt(n);i<=ed;++i){
        if(n%i==0){
            ++ans;
            int j=n/i;
            if(i!=j){
                ++ans;
            }
        }
    }
    return ans;
}

int main()
{
    //freopen("/home/lu/code/r.txt","r",stdin);
    //freopen("/home/lu/code/w.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--){
        int p;
        scanf("%d",&p);
        printf("%d\n",slove(p-1));
    }
    return 0;
}