hdu6172&&hdu6185——BM算法

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hdu6172

模板的简单应用

先根据题中的表达式求出前几项，再上BM，注意一下n的大小关系。

技术图片 
#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;
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;
    scanf("%d", &T);
    while(T--)
    
        scanf("%lld", &n);
        n--;
        /*求第n项*/
        printf("%I64d\n",linear_seq::gao(VI31,197,1255,7997,50959,324725,2069239,13185773,84023455,n-1));
View Code

hdu6185

模板的简单应用，先写个暴力程序找出前几项，可见 铺砖问题

技术图片 
#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;
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()


    while(scanf("%lld", &n) == 1)
    
        /*求第n项*/
        printf("%I64d\n",linear_seq::gao(VI1,5,11,36,95,281,781,2245,n-1));
View Code

hdu6198

模板的简单应用

先写一个爆搜找出前几项，

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1000 + 10;
bool vis[maxn];
int k;
int f[100] = 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181;

bool dfs(int n, int cnt)

    //printf("%d\n", n);
    if(n == 0 && cnt == k)  return  true;
    if(cnt == k)  return false;
    for(int i = 0;i < 20 && f[i] <= n; i++)
    
        if(dfs(n - f[i], cnt+1))  return true;  //有一个满足条件的分解即可返回
    
    return false;


int main()

    for(k = 1;k < 10;k++)
        for(int i = 0;i < 5000;i++)
        
            if(!dfs(i, 0))
            
                printf("%d %d\n",k,  i);
                break;
            
        

    return 0;

能很快找出前7项，这道题中够了。

技术图片 
#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=998244353;
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;
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()


    while(scanf("%lld", &n) == 1)
    
        /*求第n项*/
        printf("%I64d\n",linear_seq::gao(VI4, 12, 33, 88, 232, 609, 1596,n-1));

//         VI res = linear_seq::BM(VI4, 12, 33, 88, 232, 609, 1596);
//        for(int i = 1;i < res.size();i++)  printf("%lld\n", (mod-res[i]) % mod);
View Code

 

 

 

参考链接：

1. https://blog.csdn.net/Anxdada/article/details/77817850

2. https://blog.csdn.net/WilliamSun0122/article/details/77926806

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