题目1 : 骨牌覆盖问题·一 (线性递推+矩阵快速幂)

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入门的模板

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1000001;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod  1000000007

typedef struct Matrix
{
    LL m[3][3];
}mat;

mat mul(mat a,mat b)
{
    mat c;
    memset(c.m,0,sizeof(c.m));
    for(int i=0;i<1;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                c.m[i][j]+=((a.m[i][k]*b.m[k][j])%mod+mod)%mod;
    return c;
}

mat mu(mat a,mat b)
{
    mat c;
    memset(c.m,0,sizeof(c.m));
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                c.m[i][j]+=((a.m[i][k]*b.m[k][j])%mod+mod)%mod;
    return c;
}
mat matpow(mat a,int n)
{
    mat rec;
    memset(rec.m,0,sizeof(rec.m));
    rec.m[0][0]=rec.m[1][1]=rec.m[2][2]=1;
    while(n)
    {
        if(n&1)
            rec=mu(rec,a);
        a=mu(a,a);
        n>>=1;
    }
    return rec;
}

int main()
{
    //#ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    //freopen("zccccc.txt","w",stdout);
    //#endif // ONLINE_JUDGE
    int n,m,i,j,t,k;

    while(read(n))
    {
        mat a,b;
        a.m[0][0]=0;
        a.m[0][1]=1;
        b.m[0][0]=0;
        b.m[0][1]=b.m[1][0]=b.m[1][1]=1;
        //mat c=mul(a,b);
        //printf("%d %d \n%d %d\n",c.m[0][0],c.m[0][1],c.m[1][0],c.m[1][1]);
        writeln(mul(a,matpow(b,n-1)).m[0][1]%mod);
    }
    return 0;
}

题目1 : 骨牌覆盖问题·一 (线性递推+矩阵快速幂)

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原文地址：http://blog.csdn.net/u013167299/article/details/45074963