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()
{
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()
{
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;
}