矩阵快速幂模版

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输入样例是

1 1 12 也就是第12个斐波那契数列，

原理是

f(n) = |1 1| * |f(n-2)|

f(n-1)= |1,0| * |f(n-1)|

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;struct ttt{    int map1[200][200];};int n;ttt fun1(ttt &a,ttt &b){    ttt c;    int i,j,k;    for(i=1;i<=n;i++){        for(j=1;j<=n;j++){            c.map1[i][j]=0;            for(k=1;k<=n;k++)                c.map1[i][j]+=a.map1[i][k]*b.map1[k][j];        }    }    return c;}ttt fun2(ttt &a){    ttt b;    int i,j,k;    for(i=1;i<=n;i++)        for(j=1;j<=n;j++){           b.map1[i][j]=0;           for(k=1;k<=n;k++)                b.map1[i][j]+=a.map1[i][k]*a.map1[k][j];        }        return b;}ttt pow1(ttt &a,int k){    ttt b;    int i,j;    memset(b.map1,0,sizeof(b.map1));    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            if(i==j)                b.map1[i][j]=1;            while(k){                //cout << "k=" <<k <<endl;                if(k&1)                    b=fun1(b,a);                    k=k>>1;                    a=fun2(a);            }            return b;}int main(){    freopen("in.txt","r",stdin);    int i,j,k,l,f1,f2,f3,t1,t2,t3;    int r,c;    cin >>f1 >> f2>> k;    ttt a;    memset(a.map1,0,sizeof(a.map1));    n=2;    a.map1[1][1]=1;    a.map1[1][2]=1;    a.map1[2][1]=1;    a.map1[2][2]=0;    a=pow1(a,k-2);    /*for(i=1;i<=2;i++){        for(j=1;j<=2;j++)            cout << a.map1[i][j] << "  ";        cout <<endl;    }*/    cout << a.map1[1][1]*f2+a.map1[1][2]*f1 <<endl;    return 0;}

f(n)=b1*f(n-1)+a1*f(n-2)+c 输入 f1 f2 a1 b1 c1 n

例子

6 7 3 2 4 3

输出

代码:

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;struct ttt{    int map1[200][200];};int n;ttt fun1(ttt &a,ttt &b){    ttt c;    int i,j,k;    for(i=1;i<=n;i++){        for(j=1;j<=n;j++){            c.map1[i][j]=0;            for(k=1;k<=n;k++)                c.map1[i][j]+=a.map1[i][k]*b.map1[k][j];        }    }    return c;}ttt fun2(ttt &a){    ttt b;    int i,j,k;    for(i=1;i<=n;i++)        for(j=1;j<=n;j++){           b.map1[i][j]=0;           for(k=1;k<=n;k++)                b.map1[i][j]+=a.map1[i][k]*a.map1[k][j];        }        return b;}ttt pow1(ttt &a,int k){    ttt b;    int i,j;    memset(b.map1,0,sizeof(b.map1));    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            if(i==j)                b.map1[i][j]=1;            while(k){                //cout << "k=" <<k <<endl;                if(k&1)                    b=fun1(b,a);                    k=k>>1;                    a=fun2(a);            }            return b;}int main(){    freopen("in.txt","r",stdin);    int i,j,k,l,f1,f2,f3,t1,t2,t3;    int r,c;    int a1,b1,c1;    cin >>f2 >> f1>>b1>>a1>>c1>> k;    ttt a;    memset(a.map1,0,sizeof(a.map1));    n=3;    a.map1[1][1]=a1;    a.map1[1][2]=b1;    a.map1[1][3]=1;    a.map1[2][1]=1;    a.map1[2][2]=0;    a.map1[2][3]=0;    a.map1[3][1]=0;    a.map1[3][2]=0;    a.map1[3][3]=1;    a=pow1(a,k-2);    /*for(i=1;i<=3;i++){        for(j=1;j<=3;j++)            cout << a.map1[i][j] << "  ";        cout <<endl;    }*/    cout << (a.map1[1][1]*f2)+(a.map1[1][2]*f1)+a.map1[1][3]*c1 <<endl;    return 0;}

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