ICPC2019银川区域赛 D. Easy Problem

链接

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题解

首先最外层一个容斥

设

$f(n,t)=\sum_{a_1,a_2,\dots,a_n} [a_i \le \lfloor\frac{m}{t}\rfloor] \left( \prod a_i \right)^k$

答案显然等于
$\sum_{d|t}\mu(\frac{t}{d}) t^{nk} f(n,t)$

然后发现
$f(n,t) = \sum_{x=1}^{\lfloor\frac{m}{t}\rfloor} x^k f(n-1,t)\\ = f(n-1,t) \sum_{x=1}^{\lfloor\frac{m}{t}\rfloor} x^k$

预处理 $S(x) = \sum_{i=1}^{x} i^k$，则 $f(n,t) = S(\lfloor\frac{m}{t}\rfloor)^n$

最后答案就是

$\sum_{d|t}\mu(\frac{t}{d}) t^{nk} S(\lfloor\frac{m}{t}\rfloor)^n$

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 100010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
        return ans;
    }
}em;
#define mod 59964251ll
ll fact[maxn], n, k, m, d;
ll getphi(ll n)
{
    ll ret=n, i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            ret = ret/i*(i-1);
            while(n%i==0)n/=i;
        }
    }
    if(n>1)ret = ret/n*(n-1);
    return ret;
}
char s[maxn];
ll S[maxn];
int main()
{
    ll i, j, T=read(), L, ans, P=getphi(mod), t;
    em.shai(1e5+10);
    fact[0]=1; rep(i,1,1e5)fact[i]=fact[i-1]*i%mod;
    while(T--)
    {
        cl(s);
        scanf("%s%lld%lld%lld",s+1,&m,&d,&k);
        L=strlen(s+1);
        n=0;
        ans=0;
        rep(i,1,L)n=(n*10+s[i]-0x30)%P;
        rep(i,1,m)S[i]=(S[i-1]+em.fastpow(i,k,mod))%mod;
        rep(t,d,m)
        {
            if(t%d)continue;
            if(em.mu[t/d]==0)continue;
            ll tmp = em.mu[t/d] * em.fastpow(t,n*k%P+P,mod) %mod * em.fastpow(S[m/t],n+P,mod);
            ( ans += tmp) %=mod;
        }
        printf("%lld\n",(ans+mod)%mod);
    }
    return 0;
}