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  • ProjectEuler654

    我,ycl:BM是什么早就忘了!

    毕老爷:那你们可以做一做这道题练练BM板子啊。

    传送门

     1 //Achen
     2 #include<bits/stdc++.h>
     3 #define For(i,a,b) for(int i=(a);i<=(b);i++)
     4 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
     5 #define Formylove return 0
     6 const int N=1e5+7,p=1e9+7;
     7 typedef long long LL;
     8 typedef double db;
     9 using namespace std;
    10 int n,m;
    11 LL f[15007][5007];
    12 
    13 template<typename T> void read(T &x) {
    14     char ch=getchar(); T f=1; x=0;
    15     while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    16     if(ch=='-') f=-1,ch=getchar();
    17     for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
    18 }
    19 
    20 int main() {
    21     freopen("1.in","r",stdin);
    22     //freopen("biao.in","w",stdout);
    23     read(m); read(n);
    24     For(i,1,m) f[1][i]=1;
    25     For(i,1,n) {
    26         if(i>1) For(j,1,m) f[i][j]=f[i-1][m-j];
    27         For(j,1,m) f[i][j]=(f[i][j]+f[i][j-1])%p;
    28     }
    29     printf("%lld
    ",f[14555][m]);
    30     //printf("%d
    ",n);
    31     //For(i,1,n) printf("%lld ",f[i][m]);
    32     //printf("%lld
    ",f[n][m]);
    33     Formylove;
    34 }
    n^2暴力
    //Achen
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<vector>
    #include<cstdio>
    #include<queue>
    #include<cmath>
    #include<set>
    #include<map>
    #define Formylove return 0
    #define For(i,a,b) for(int i=(a);i<=(b);i++)
    #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
    const int N=15007,p=1e9+7;
    typedef long long LL;
    typedef double db;
    using namespace std;
    int n,cnt,fail[N];
    vector<LL>f[N];
    LL a[N],delta[N];
    
    template<typename T>void read(T &x)  {
        char ch=getchar(); x=0; T f=1;
        while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
        if(ch=='-') f=-1,ch=getchar();
        for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
    }
    
    LL ksm(LL a,LL b) {
        LL rs=1,bs=a%p;
        while(b) {
            if(b&1) rs=rs*bs%p;
            bs=bs*bs%p;
            b>>=1;
        }
        return rs;
    }
    
    #define ANS
    int main() {
    #ifdef ANS
        freopen("biao.in","r",stdin);
        freopen("shizi.out","w",stdout);
    #endif
        read(n);
        For(i,1,n) read(a[i]);
        For(i,1,n) {
            LL tp=0; int up=f[cnt].size();
            For(j,0,up-1) tp=(tp+f[cnt][j]*a[i-j-1]%p)%p;
            if(tp==a[i]) continue;
            delta[i]=(a[i]-tp+p)%p;
            fail[cnt]=i;
            ++cnt;
            if(cnt==1) {
                f[cnt].resize(i);
                continue;
            }
            LL mul=delta[i]*ksm(delta[fail[cnt-2]],p-2)%p,fmul=(p-mul)%p;
            f[cnt].resize(i-fail[cnt-2]-1);
            f[cnt].push_back(mul);
            up=f[cnt-2].size();
            For(j,0,up-1) f[cnt].push_back(fmul*f[cnt-2][j]%p);
            up=f[cnt-1].size();
            if(f[cnt].size()<f[cnt-1].size()) f[cnt].resize(up);
            For(j,0,up-1) f[cnt][j]=(f[cnt][j]+f[cnt-1][j])%p; 
        }
        int up=f[cnt].size();
        printf("%d
    ",up);
        For(i,0,up-1) printf("%lld ",f[cnt][i]);
        Formylove;
    }
    BM出递推
     1 //Achen
     2 #include<bits/stdc++.h>
     3 #define For(i,a,b) for(int i=(a);i<=(b);i++)
     4 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
     5 #define Formylove return 0
     6 const int N=1e5+7,p=1e9+7;
     7 typedef long long LL;
     8 typedef double db;
     9 using namespace std;
    10 int cnt;
    11 LL A[N],f[N],n;
    12 
    13 template<typename T> void read(T &x) {
    14     char ch=getchar(); T f=1; x=0;
    15     while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    16     if(ch=='-') f=-1,ch=getchar();
    17     for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
    18 }
    19 
    20 LL rs[N],bs[N],tp[N];
    21 void cheng(LL a[],LL b[]) {
    22     For(i,0,cnt*2) tp[i]=0;
    23     For(i,0,cnt-1) For(j,0,cnt-1)
    24         (tp[i+j]+=a[i]*b[j])%=p;
    25     Rep(i,cnt*2-2,cnt) if(tp[i])
    26         For(j,0,cnt-1) 
    27             (tp[i-j-1]+=A[j]*tp[i]%p)%=p;
    28     For(i,0,cnt-1) a[i]=tp[i];
    29 }
    30 
    31 void ksm(LL b) {
    32     while(b) {
    33         if(b&1) cheng(rs,bs);
    34         cheng(bs,bs);
    35         b>>=1;
    36     }
    37 }
    38 
    39 int main() {
    40     freopen("shizi.out","r",stdin);
    41     //freopen("shizi.out","w",stdout);
    42     read(cnt);
    43     For(i,0,cnt-1) read(A[i]);
    44     read(n);
    45     rs[0]=1; bs[1]=1; ksm(n-1);
    46     For(i,1,cnt) read(f[i]);
    47     LL ans=0;
    48     For(i,0,cnt-1) ans=(ans+f[i+1]*rs[i]%p)%p;
    49     printf("%lld
    ",ans);
    50     Formylove;
    51 }
    多项式取模
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  • 原文地址:https://www.cnblogs.com/Achenchen/p/10518986.html
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