HDU6108 小C的倍数问题 【脑洞】

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证明:p进制下,当且仅当,x每位之和为B的倍数时,x%B=0

设x=k0+k1p+k2p2+...+knpn
若(k0+k1+...+kn)%B=∑ni=0ki%B=0
又因为x=∑ni=0ki+∑ni=1((pi−1)∗ki)
所以当∑ni=1((pi−1)∗ki)%B=0时,x%B=0
而pi−1=(p−1)∗(pi−1+pi−2+...+1)−>(pi−1)%(p−1)=0   
所以当B=p−1或p−1的因子时,有:当且仅当∑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;}