nyist 301 递推求值（矩阵快速幂）

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题目地址：http://acm.nyist.net/JudgeOnline/problem.php?pid=301

思路：注意矩阵乘法不满足交换律，所以要注意初始矩阵在右边，要进行快速幂的矩阵在左边

左边的矩阵为右边的矩阵为

b a c f2 0 0

1 0 0 f1 0 0

0 0 1 1 0 0

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long ll;const int maxn = 40000;const int mod = 1000007;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;struct node{    ll a[3][3];};node matrixmul(node a,node b){    node c;    for(int i=0; i<3; i++)    {        for(int j=0; j<3; j++)        {            c.a[i][j] = 0;            for(int k=0; k<3; k++)            {                c.a[i][j] += a.a[i][k] * b.a[k][j];                c.a[i][j] %= mod;            }        }    }    return c;}node quickmod(node s,int k){    node ans;    memset(ans.a,0,sizeof(s.a));    ans.a[0][0]=ans.a[1][1]=ans.a[2][2]=1;//初始化为单位矩阵，值不变    while(k)    {        if(k & 1)            ans = matrixmul(ans,s);        k = k >> 1;        s = matrixmul(s,s);    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int f1,f2,d,e,f,n;        scanf("%d%d%d%d%d%d",&f1,&f2,&d,&e,&f,&n);        if(n == 1)        {             printf("%d\n",(f1+mod)%mod);             continue;        }        if(n == 2)        {            printf("%d\n",(f2+mod)%mod);            continue;        }        node sa;        memset(sa.a,0,sizeof(sa.a));        sa.a[0][0]=e;        sa.a[0][1]=d;        sa.a[0][2]=f;        sa.a[1][0]=1;        sa.a[2][2]=1;        sa = quickmod(sa,n-2);        node sb;        memset(sb.a,0,sizeof(sb.a));        sb.a[0][0]=f2;        sb.a[1][0]=f1;        sb.a[2][0]=1;        sb = matrixmul(sa,sb);//注意顺序        ll ans = (sb.a[0][0]+mod)%mod;        printf("%I64d\n",ans);    }    return 0;}

大神地址：http://www.lxway.com/146981016.htm

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