Matrix Fast Power POJ 3070 Fibonacci

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1 /*2 Matrix Fast Power: To find the Fibonacci number of the nth term, transpose the matrix are given, set a template can be. It's high efficiency.3 */4#include <cstdio>5#include <algorithm>6#include <cstring>7#include <cmath>8 using namespacestd;9 Ten Const intMAXN = 1e3 +Ten; One Const intINF =0x3f3f3f3f; A Const intMOD =10000; - structMat { -     intm[2][2]; the }; -  - Mat Multi_mod (Mat A, Mat b) { -MAT ret; memset (RET.M,0,sizeof(RET.M)); +      for(intI=0; i<2; ++i) { -          for(intj=0; j<2; ++j) { +              for(intk=0; k<2; ++k) { ARET.M[I][J] = (Ret.m[i][j] + a.m[i][k] * b.m[k][j])%MOD; at             } -         } -     } -     returnret; - } -  in intMatrix_pow_mod (Mat x,intN) { -MAT ret; ret.m[0][0] = ret.m[1][1] =1; ret.m[0][1] = ret.m[1][0] =0; to      while(n) { +         if(N &1) ret =Multi_mod (ret, x); -x =multi_mod (x, x); theN >>=1; *     } $     returnret.m[0][1];Panax Notoginseng } -  the intMainvoid)  {//POJ 3070 Fibonacci +     intN; A      while(SCANF ("%d", &n) = =1)   { the         if(n = =-1) Break; + Mat x; -x.m[0][0] = x.m[0][1] = x.m[1][0] =1; x.m[1][1] =0; $printf ("%d\n", Matrix_pow_mod (x, N)); $     } -  -     return 0; the}

Matrix Fast Power POJ 3070 Fibonacci