ZOJ 2317 Nice Patterns Strike Back（矩阵快速幂）

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2317     

题意：给你两种颜色，黑色和白色，填充n*m的方格，每个格子一种颜色，但要保证在每个2*2的方格内不能出现同一中颜色

题解：枚举当前行和上一行的所有合法状态，矩阵快速幂

  假如：2 * 2的方格，合法状态为1

         
 {0 1 1 1}
 {1 1 1 1}

sum+=  {1,1,1,1}  * {1 1 1 1}   

 {1 1 1 0}

可以想想。。。。。。。

/*
* this code is made by fenger
* Problem: 1214
* Verdict: Accepted
* Submission Date: 2014-10-19 16:46:55
* Time: 212MS
* Memory: 1864KB
*/
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int m;
int p[35][35];
int cnt,mod;
struct Z{
int m[33][33];
Z(){
memset(m,0,sizeof(m));
}
void init(){
for(int i = 0;i < cnt;i++)
m[i][i] = 1;
}
};
string solve(string a)//大数除法
{
string s;
int l = a.size();
int c = 1,num = 0,ans = a[0] - '0';
if(ans >= 2) {c = 0;ans = 0;s="";}
for(int i = c;i < l;i++)
{
ans = ans*10 + a[i] - '0';
s += ans / 2 + '0';
ans %= 2;
}
if(s == "") s = "0";
return s;
}
int Pow(int a,string x){//2^x%mod
int res = 1;

while(x[0] != '0'){
if((x[x.size()-1] -'0')&1) res = res * a % mod;
a = a * a % mod;
x = solve(x);
}
return res;
}
Z operator * (Z a , Z b){//矩阵乘法
Z c;
for(int i = 0;i < cnt;i++)
for(int k = 0;k < cnt;k++)
if(a.m[i][k])
for(int j = 0;j < cnt;j++)
c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]) % mod;
return c;
}
Z Pow1(string x,Z a){//矩阵快速幂
Z ret;
ret.init();
while(x[0]!='0'){
if((x[x.size()-1] - '0')&1) ret = ret * a;
a = a * a;
x = solve(x);
}
return ret;
}
string mul(string a)//大数减法
{
int l = a.size();
if(a[l-1] - '0' >= 1){
a[l-1] -= 1;
return a;
}
else{
string s = "";
a[l-1] += 9;
for(int i = l - 2;i >= 0;i--)
if(a[i] == '0')  a[i] = '9';
else {a[i] -= 1;break;}
if(a[0] == '0'){
for(int i = 1;i < l;i++) s += a[i];
return s;
}
else {return a;}
}
}
void dfs(int cur , int now ,int pre){//枚举当前行，与上一行的合法状态
if(cur > m) return;
if(cur == m)
{
p[pre][now] = 1;
return ;
}
if(cur == 0){
dfs(cur+1,(now<<1)|1,(pre<<1)|1);
dfs(cur+1,(now<<1)|1,(pre<<1));
dfs(cur+1,(now<<1),(pre<<1));
dfs(cur+1,(now<<1),(pre<<1)|1);
}
else{
int k1 = now&1,k2 = pre&1;
if(k1 == k2 && k1){

dfs(cur+1,(now<<1)|1,(pre<<1));
dfs(cur+1,(now<<1),(pre<<1)|1);
dfs(cur+1,(now<<1),(pre<<1));

}
else if(k1 == k2&&(!k1)){
dfs(cur+1,(now<<1),(pre<<1)|1);
dfs(cur+1,(now<<1)|1,(pre<<1));
dfs(cur+1,(now<<1)|1,(pre<<1)|1);
}
else if(k1 != k2){
dfs(cur+1,(now<<1)|1,(pre<<1)|1);
dfs(cur+1,(now<<1)|1,(pre<<1));
dfs(cur+1,(now<<1),(pre<<1));
dfs(cur+1,(now<<1),(pre<<1)|1);
}
}
}
int main(){
string str;
while(cin >> str >> m >> mod){
int l = str.size();
if(m == 1){
int w = Pow(2,str);
cout << w << endl;
}
else if(l == 1 && str[0] == '1'){
int w = pow(2,m);
cout << (w%mod) << endl;
}
else{
cnt = (1 << m);
memset(p,0,sizeof(p));
dfs(0,0,0);
Z s;
for(int i = 0;i < cnt;i++)
for(int j = 0;j < cnt;j++)
s.m[i][j] = p[i][j];
str = mul(str);
s = Pow1(str,s);
int ans = 0;
for(int i = 0;i < cnt;i++)
for(int j = 0;j < cnt;j++)
ans = (ans + s.m[i][j])%mod;
cout << ans << endl;
}
}
}