来源自http://blog.csdn.net/chenguolinblog/article/details/10309423
hdu 1575
Tr A
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5587 Accepted Submission(s): 4200
Problem Description
A为一个方阵，则Tr A表示A的迹（就是主对角线上各项的和），现要求Tr(A^k)%9973。
Input
数据的第一行是一个T，表示有T组数据。
每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行，每行有n个数据，每个数据的范围是[0,9]，表示方阵A的内容。
Output
对应每组数据，输出Tr(A^k)%9973。
Sample Input
2 2 2 1 0 0 1 3 99999999 1 2 3 4 5 6 7 8 9
Sample Output
2 2686
裸矩阵快速幂。
1 #include<bits/stdc++.h>
2 #define clr(x) memset(x,0,sizeof(x))
3 #define LL long long
4 #define mod 9973
5 using namespace std;
6 typedef vector<LL> vec;
7 typedef vector<vec> mat;
8 mat mul(const mat &a,const mat &b)
9 {
10 int row=a.size();
11 int col=b[0].size();
12 int mid=b.size();
13 mat c(row,vec(col));
14 for(int i=0;i<row;i++)
15 for(int j=0;j<col;j++)
16 for(int k=0;k<mid;k++)
17 c[i][j]=(c[i][j]+a[i][k]*b[k][j]%mod)%mod;
18 return c;
19 }
20 mat quick_pow(mat a,LL n)
21 {
22 int len=a.size();
23 mat res(len,vec(len));
24 for(int i=0;i<len;i++)
25 res[i][i]=1;
26 while(n)
27 {
28 if(n&1)
29 res=mul(res,a);
30 a=mul(a,a);
31 n>>=1;
32 }
33 return res;
34 }
35 int main()
36 {
37 int T,n;
38 LL k,ansed;
39 scanf("%d",&T);
40 while(T--)
41 {
42 scanf("%d%lld",&n,&k);
43 mat ans(n,vec(n));
44 for(int i=0;i<n;i++)
45 for(int j=0;j<n;j++)
46 scanf("%lld",&ans[i][j]);
47 ans=quick_pow(ans,k);
48 ansed=0;
49 for(int i=0;i<n;i++)
50 ansed=(ans[i][i]+ansed)%mod;
51 printf("%lld\n",ansed);
52 }
53 return 0;
54 }
hdu 1757
A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5264 Accepted Submission(s): 3200
Problem Description
Lele now is thinking about a simple function f(x).
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
Input
The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
Output
For each case, output f(k) % m in one line.
Sample Input
10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0
Sample Output
45 104
裸矩阵快速幂。
1 #include<bits/stdc++.h>
2 #define clr(x) memset(x,0,sizeof(x))
3 #define LL long long
4 using namespace std;
5 typedef vector<LL> vec;
6 typedef vector<vec> mat;
7 int a[20];
8 LL m,k,p,n;
9 mat mul(const mat &a,const mat &b,LL mod)
10 {
11 mat c(a.size(),vec(b[0].size()));
12 for(int i=0;i<a.size();i++)
13 for(int j=0;j<b[0].size();j++)
14 for(int k=0;k<b.size();k++)
15 {
16 c[i][j]=(c[i][j]%mod+a[i][k]*b[k][j]%mod)%mod;
17 }
18 return c;
19 }
20 mat quick_pow(mat a,LL n,LL mod)
21 {
22 int sized=a.size();
23 mat res(sized,vec(sized));
24 for(int i=0;i<10;i++)
25 res[i][i]=1;
26 while(n)
27 {
28 if(n&1)
29 {
30 res=mul(res,a,mod);
31 }
32 n>>=1;
33 a=mul(a,a,mod);
34 }
35 return res;
36 }
37 int main()
38 {
39 while(scanf("%lld%lld",&n,&m)!=EOF)
40 {
41 for(int i=0;i<10;i++)
42 scanf("%lld",&a[i]);
43 if(n<10)
44 {
45 printf("%lld\n",n%m);
46 continue;
47 }
48 mat multi(10,vec(10));
49 mat ans(1,vec(10));
50 for(int i=0;i<10;i++)
51 {
52 ans[0][i]=i;
53 multi[i][9]=a[9-i];
54 }
55 for(int i=0;i<9;i++)
56 multi[i+1][i]=1;
57 multi=quick_pow(multi,n-9,m);
58 ans=mul(ans,multi,m);
59 printf("%lld\n",ans[0][9]);
60 }
61 return 0;
62 }