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Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4203 Accepted Submission(s):
3142
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define mod 9973 5 using namespace std; 6 struct mat 7 { 8 int m[10][10]; 9 }; 10 int n; 11 mat mul(mat a,mat b) 12 { 13 mat c; 14 int i,j,k; 15 memset(c.m,0,sizeof(c.m)); 16 for(i=0; i<n; i++) 17 for(j=0; j<n; j++) 18 { 19 for(k=0; k<n; k++) 20 c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod; 21 c.m[i][j]%=mod; 22 } 23 return c; 24 } 25 26 mat product(mat a,int k) 27 { 28 if(k==1) return a; 29 else if(k&1) 30 return mul(product(a,k-1),a); 31 else 32 return product(mul(a,a),k/2); 33 } 34 35 int main() 36 { 37 int T,i,j,k; 38 mat a,b; 39 scanf("%d",&T); 40 while(T--) 41 { 42 scanf("%d%d",&n,&k); 43 for(i=0; i<n; i++) 44 for(j=0; j<n; j++) 45 scanf("%d",&a.m[i][j]); 46 b=product(a,k); 47 int ans=0; 48 for(i=0; i<n; i++) 49 ans=(ans+b.m[i][i])%mod; 50 printf("%d\n",ans); 51 } 52 return 0; 53 }
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原文地址:http://www.cnblogs.com/pshw/p/5545931.html