HDU.2256 Problem of Precision (矩阵快速幂)

HDU.2256 Problem of Precision (矩阵快速幂)

题意分析

代码总览

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#define INF 0x3f3f3f3f
#define nmax 200
#define MEM(x) memset(x,0,sizeof(x))
using namespace std;
const int Dmax = 11;
int N = 2;
int MOD;
typedef struct{
    int matrix[Dmax][Dmax];
    void init()//初始化为单位矩阵
    {
        memset(matrix,0,sizeof(matrix));
        for(int i = 0; i<Dmax;++i) matrix[i][i] = 1;
    }
}MAT;

MAT ADD(MAT a, MAT b)
{
    for(int i = 0; i<N;++i){
        for(int j = 0;j<N;++j){
            a.matrix[i][j] +=b.matrix[i][j];
            a.matrix[i][j] %= MOD;
        }
    }
    return a;
}
MAT MUL(MAT a, MAT b)
{
    MAT ans;
    for(int i = 0; i<N;++i){
        for(int j = 0; j<N;++j){
            ans.matrix[i][j] = 0;
            for(int k = 0; k<N;++k){
                ans.matrix[i][j] += ( (a.matrix[i][k]) % MOD * (b.matrix[k][j]) % MOD) % MOD;
            }
            ans.matrix[i][j] %= MOD;
        }
    }
    return ans;
}
MAT POW(MAT a, int t)
{
    MAT ans; ans.init();
    while(t){
        if(t&1) ans = MUL(ans,a);
        t>>=1;
        a = MUL(a,a);
    }
    return ans;
}
void OUT(MAT a)
{
    for(int i = 0; i<N;++i){
        for(int j =  0; j<N;++j){
            printf("%5d",a.matrix[i][j]);
        }
        printf("
");
    }
}
void IN(MAT & a,MAT & ini)
{
    memset(a.matrix,0,sizeof(a.matrix));
    memset(ini.matrix,0,sizeof(ini.matrix));
    a.matrix[0][0] = 5;
    a.matrix[0][1] = 12;
    a.matrix[1][0] = 2;
    a.matrix[1][1] = 5;
    ini.matrix[0][0] = 5;
    ini.matrix[1][0] = 2;
}
void CAL(MAT a)
{
    printf("%d
",(2* a.matrix[0][0] -1) % MOD);
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n;MOD = 1024;
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        MAT temp,ini;
        IN(temp,ini);
        temp = POW(temp,n-1);
        temp = MUL(temp,ini);
        CAL(temp);
    }
    return 0;
}