Infiniti

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#include <bits/stdc++.h>

using namespace std;
#define rep(i,a,n) for (long long i=a;i<n;i++)
#define per(i,a,n) for (long long i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((long long)(x).size())
typedef vector<long long> VI;
typedef long long ll;
typedef pair<long long,long long> PII;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head

long long _,n,c,k,a[11115],fib[11020]={0,1};
namespace linear_seq
{
    const long long N=10010;
    ll res[N],base[N],_c[N],_md[N];
    vector<long long> Md;
    void mul(ll *a,ll *b,long long k)
    {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k)
            _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (long long i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    long long solve(ll n,VI a,VI b)
    { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(b));
        ll ans=0,pnt=0;
        long long k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (long long p=pnt;p>=0;p--)
        {
            mul(res,res,k);
            if ((n>>p)&1)
            {
                for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s)
    {
        VI C(1,1),B(1,1);
        long long L=0,m=1,b=1;
        rep(n,0,SZ(s))
        {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n)
            {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            }
            else
            {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    long long gao(VI a,ll n)
    {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};

int main()
{
    int T,ide=0;
    scanf("%d",&T);
    while(T--){
    scanf("%lld%lld%lld",&n,&c,&k);
    for(int i=2;i<10000;i++)fib[i]=(fib[i-1]+fib[i-2])%mod;
    for(int i=0;i<=101;i++)
    a[i]=powmod(fib[i*c],k);
    printf("Case %d: ",++ide);
    VI qq;qq.clear();
    for(int i=0;i<=100;i++){
    if(i>0)(a[i]+=a[i-1])%=mod;
    qq.push_back(a[i]);
    }
    printf("%lld\n",linear_seq::gao(qq,n));
    }
    return 0;
}

  

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const LL MAXN = 111;
long long aa,bb;
LL K,C[MAXN], M[MAXN],x,y,m;
LL gcd(LL a, LL b)
{
    return b == 0 ? a : gcd(b, a % b);
}
LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    LL r = exgcd(b, a % b, x, y), tmp;
    tmp = x;
    x = y;
    y = tmp - (a / b) * y;
    return r;
}
LL inv(LL a, LL b)
{
    LL r = exgcd(a, b, x, y);
    while (x < 0)x+=b;
    return x;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
    K=3;
    for (LL i = 1; i <= K; i++)
    scanf("%lld", &M[i]);
    for (LL i = 1; i <= K; i++)
    scanf("%lld", &C[i]);
    bool flag = 1;
    for (LL i = 2; i <= K; i++)
    {
        LL M1 = M[i - 1], M2 = M[i], C2 = C[i], C1 = C[i - 1], T = gcd(M1, M2);
        if ((C2 - C1) % T != 0)
        {
            flag = 0;
            break;
        }
        M[i] = (M1 * M2) / T;
        C[i] = ( inv( M1 / T, M2 / T ) * (C2 - C1) / T ) % (M2 / T) * M1 + C1;
        C[i] = (C[i] % M[i] + M[i]) % M[i];
    }
    aa=C[K];
    for(int i=0;;i++){
        LL temp=1ll*i*i*i;
        temp%=M[K];
        if(temp==aa){
            printf("%d\n",i);
            break;
        }
    }
    }
    return 0;
}

  

posted on 2019-09-20 18:23  自由缚  阅读(143)  评论(0编辑  收藏  举报