今日SGU 5.7

SGU 169

题意:求k位数里面有多少个是完美数,完美数的定义就是n是好数,n+1也是好数,那么n就是完美数,好数就是n%p(n)==0&&p(n)!=0,p(n)=a1*...*an

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#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int maxn = 1e5+5;
const double eps = 1e-6;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int bit(ll n){
    int ret = 0;
    while(n) n/=10,ret++;
    return ret;
}
//ll p(ll n){
//    ll ret = 1;
////    dd(n)
//    while(n) ret*=(n%10),n/=10;
////    de(ret)
//    return ret;
//}
//bool check(ll n){
//    ll tmp = p(n);
//    if(tmp==0) return false;
//    if(n%tmp==0) return true;
//    return false;
//}
//bool ok(ll n){
//    if(check(n)&&check(n+1)) return true;
//    return false;
//}
int main(){
//    rep(i,1,1e9+1) {
//        if(ok(i)) {
//            dd(bit(i))de(i)
//        }
//    }
    qc;
    int k;
    cin>>k;--k;
    if(!k) printf("8");
    else {
        if(k%6==0) printf("4");
        else if(k%3==0) printf("3");
        else printf("1");
    }
    return 0;
}
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转载自www.cnblogs.com/chinacwj/p/9005397.html
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