Matrix Power Series(行列高速べき乗)
3210 ワード
行列の高速べき乗:http://www.cnblogs.com/kuangbin/archive/2012/08/17/2643347.html
Matrix Power Series
Time Limit:3000 MS
メモリLimit:13131313722 K
Total Submissions:16341
Acceepted:6966
Description
Given a n× n matix A and a positive integer k,find the sumS=A 2+A 3+…+Ak.
Input
The input contains exactly one test case.The first line of input contains three positive integers n(n≦30)、k(k≦109)and m(m<104).The n follown line eas each containing n nonegative integers belogint belogint
Output
Output the elemens of S modulo m in the same way as A is given.
Sample Input
二次二次マトリックス、具体的にコードを見ましょう。
ACコード:
Matrix Power Series
Time Limit:3000 MS
メモリLimit:13131313722 K
Total Submissions:16341
Acceepted:6966
Description
Given a n× n matix A and a positive integer k,find the sumS=A 2+A 3+…+Ak.
Input
The input contains exactly one test case.The first line of input contains three positive integers n(n≦30)、k(k≦109)and m(m<104).The n follown line eas each containing n nonegative integers belogint belogint
Output
Output the elemens of S modulo m in the same way as A is given.
Sample Input
2 2 4
0 1
1 1
Sample Output1 2
2 3
簡単で分かりやすいです。二次二次マトリックス、具体的にコードを見ましょう。
ACコード:
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-9
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int M = 1e5 + 100;
const int INF = 0x3f3f3f3f;
int n,mod,k;
struct Mac {
int c[33][33]; //
void unit1() { //
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
scanf("%d",&c[i][j]);
}
}
}
void unit2() { //
c[0][0] = c[1][1] = 1;
c[1][0] = c[0][1] = 0;
}
};
Mac tmp,sum; // ,
Mac mul(Mac a,Mac b) { //
Mac p;
memset(p.c,0,sizeof(p.c));
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(a.c[i][j]) {
for(int k = 0; k < n; k++) {
p.c[i][k] += (a.c[i][j] * b.c[j][k]) % mod;
p.c[i][k] %= mod;
}
}
}
}
return p;
}
Mac add(Mac a,Mac b) { //
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
a.c[i][j] += b.c[i][j] % mod;
a.c[i][j] %= mod;
}
}
return a;
}
Mac quickmod(Mac a,int n) { //
Mac p;
p.unit2();
while(n) {
if(n & 1) p = mul(p,a);
a = mul(a,a);
n >>= 1;
}
return p;
}
Mac binary_matrix1(int k) { //
Mac res;
if(k == 1) {
return tmp;
}
res = binary_matrix1(k / 2);
res = mul(res,res);
if(k & 1) res = mul(res,tmp);
return res;
}
Mac binary_matrix2(int k){ // A1+A2+A3...;
if(k == 1){
return tmp;
}
Mac res = binary_matrix2(k / 2);
Mac tp = binary_matrix1(k / 2);
tp = mul(tp,res);
sum = add(sum,tp);
if(k & 1) sum = add(sum,binary_matrix1(k));
return sum;
}
int main() {
cin>>n>>k>>mod;
tmp.unit1();
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
sum.c[i][j] = tmp.c[i][j];
}
}
Mac res = binary_matrix2(k);
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
printf(j == n - 1 ? "%d
" : "%d ",res.c[i][j]);
}
}
return 0;
}