Mathematical formula + matrix fast Power -2013 ACM/ICPC Asia Regional Changsha Online H question __ Maths

Topic Link:

Http://acm.zju.edu.cn/changsha/showContestProblem.do?problemId=19

Topic Meaning:

The subject is very obscure. In the K-dimensional space, the start length is L, and the operation becomes l+√ (l* (l+1)). Ask the last%k the number of integers left.

Ideas for solving problems:

Mathematical formula + matrix fast power.

Because it is in the K-dimensional space, it is also required (l+√ (l* (L-1)) ^k The value of the%k after the rounding down.

This question and this year, the Changsha Invitational contest a question, almost, but the problem is to take the whole up, this problem is to take the whole, minus one on the line.

Thinking reference Hdu 4565

Code:

#include <iostream> #include <cmath> #include <cstdio> #include <cstdlib> #include <string > #include <cstring> #include <algorithm> #include <vector> #include <map> #include <set > #include <stack> #include <list> #include <queue> #include <ctime> #define EPS 1e-6 #define INF 0x3f3f3f3f #define PI acos ( -1.0) #define LL long #define Lson l,m, (rt<<1) #define Rson m+1,r, (rt<<1) |

1 #pragma COMMENT (linker, "/stack:1024000000,1024000000") using namespace std;

#define MAXN 3 ll m,a,b;
    struct Mar {int r,c;
    ll SA[MAXN][MAXN];
        void init (int a,int b) {r=a,c=b;
    memset (sa,0,sizeof (SA));
}
};
    Mar operator * (struct Mar & a,struct Mar & B) {Mar C;

    C.init (A.R,B.C);
                for (int k=1;k<=a.c;k++) {for (int i=1;i<=c.r;i++) {if (!a.sa[i][k))
            Continue
           for (int j=1;j<=c.c;j++) {     if (!b.sa[k][j]) continue;
            C.sa[i][j]= (C.sa[i][j]+a.sa[i][k]*b.sa[k][j])%m;
}} return C;

Mar Ba[35],ans;
    void Init () {ba[0].init (2,2);
    ba[0].sa[1][1]= (2*a)%m,ba[0].sa[1][2]= ((b-a*a)%m+m)%m;


    ba[0].sa[2][1]=1,ba[0].sa[2][2]=0;
    for (int i=1;i<32;i++)//2^i {ba[i]=ba[i-1]*ba[i-1];

   int main () {ll l,k;
       while (~SCANF ("%lld%lld", &k,&l)) {a=l,b=l* (L-1);
       M=k;
       Init ();
       Ans.init (2,1);

       ans.sa[1][1]= (2*a)%m,ans.sa[2][1]=2;
           for (int i=0;i<32&&k;i++) {if (k&1) {Ans=ba[i]*ans;
       } k>>=1;
   printf ("%lld\n", (ans.sa[2][1]-1+m)%m);
return 0;

 }