aaaaaaa……aaa(n个)%p的值 (矩阵快速幂)

觉得还是有必要发日志的……否则就在题海中迷失了自己的脚步咯

题目如标题，下面是程序……不要问我矩阵是怎么想的……

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12 const int MAX=5;
13 LL a,n,p;
14 struct Mat{
15     LL x,y;
16     LL mat[MAX][MAX];
17     Mat (){
18         x=y=0;
19         memset(mat,0,sizeof(mat));
20     }
21     Mat operator * (const Mat &cc) {
22         int i,j,k;
23         Mat zt;
24         zt.x=x,zt.y=cc.y;
25         for (i=1;i<=zt.x;i++)
26          for (j=1;j<=cc.y;j++)
27           for (k=1;k<=cc.x;k++)
28            zt.mat[i][j]=(zt.mat[i][j]+mat[i][k]*cc.mat[k][j])%p;
29         return zt;
30     }
31 }m1,m2;
32 void init(){
33     int i,j;
34     scanf("%lld%lld%lld",&a,&n,&p);
35     m1.x=2,m1.y=1;
36     m1.mat[1][1]=m1.mat[2][1]=a;
37     m2.x=2,m2.y=2;
38     m2.mat[1][1]=10,m2.mat[1][2]=1;
39     m2.mat[2][1]=0,m2.mat[2][2]=1;
40 }
41 Mat ksm(Mat zt,LL k){
42     Mat an;
43     LL i,j;
44     an.x=zt.x,an.y=zt.y;
45     for (i=1;i<=an.x;i++)
46      for (j=1;j<=an.y;j++)
47       an.mat[i][j]=(i==j);
48     while (k)
49     {if (k%2==1)
50       an=an*zt;
51      zt=zt*zt;
52      k/=2;
53     }
54     return an;
55 }
56 int main(){
57     freopen ("ksm.in","r",stdin);
58     freopen ("ksm.out","w",stdout);
59     int i,j;
60     init();
61     m1=ksm(m2,n-1)*m1;
62     printf("%lld",m1.mat[1][1]);
63     return 0;
64 }

posted on 2016-10-10 21:01 珍珠鸟阅读(246) 评论(0) 编辑收藏举报