二分求和
或者矩陣套矩陣
|A E| ^n  = |A^n E+...+A^(n-1)|
|0 E|       |0   E            |
https://cn.vjudge.net/solution/9721123 
這個代碼我卻不懂了= =



#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <queue>
#include <math.h>
#include <iostream>
using namespace std;
#define LL long long
const
 
 int maxn=(int)1e5+5;
LL MOD=8000;
int n;
struct mat{
    LL a[2][2];
    mat(){
        memset(a,0,sizeof a);
    }
    mat operator *(const mat &q){
        mat t;
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++ ){
            LL w=0;
            for(int k=0;k<n;k++)w=(w+a[i][k]*q.a[k][j])%MOD;
            t.a[i][j]
 
=w;
        }
        return t;
    }
    mat operator +(const mat& q){
        mat t;memset(t.a,0,sizeof t.a);
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++){
            t.a[i][j]=(a[i][j]+q.a[i][j])%MOD;
        }
        return t;
    }
    void initE(){
         a[1][0]=a[0][1]=0
 
;a[1][1]=a[0][0]=1;
    }
    void initFib(){
        for(int i=0;i<n;i++)for(int j=0;j<n;j++)a[i][j]=1;a[1][1]=0;
    }
    void in(){
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)scanf("%lld",&a[i][j]);
    }
    void print(){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++)printf("%lld ",a[i][j]);printf("\n");
        }
    }
};
LL k,b,N;
mat quickmatpower(mat a ,LL k){
    mat res;res.initE();
    while(k>0){
        if(k&1)res=res*a;
        a=a*a;
        k>>=1LL;
    }
    return res;
}
LL quickpower(LL a,LL k){
    LL res=1;
    while(k>0){
        if(k&1)res=res*a;
        a=a*a;
        k>>=1LL;
    }
    return res;
}
mat calsum(mat a,LL k){
    mat E;E.initE();
    if(k==1) return a;
    if(k%2==0){
        return calsum(a,k/2)*(E+quickmatpower(a,k/2));
    }
    return calsum(a,k/2)*(E+quickmatpower(a,k/2))+quickmatpower(a,k);
}
LL work(){
    mat f;f.initFib();mat E;E.initE();
    return (quickmatpower(f,b)*((E+calsum(quickmatpower(f,k),N-1)))).a[0][1];
}
int main()
{
    #ifdef shuaishuai
    freopen("C:\\Users\\hasee\\Desktop\\a.txt","r",stdin);
//   freopen("C:\\Users\\hasee\\Desktop\\b.txt","w",stdout);
#endif
    n=2;
    while(~scanf("%lld%lld%lld%lld",&k,&b,&N,&MOD)){
        printf("%lld\n",work());
    }
    return 0;
}

Gauss Fibonacci HDU - 1588 等比矩陣列求和