hdu 2276 & nyoj 300

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

There are n lights in a circle numbered from 1 to n. The left of light 1 is light n, and the left of light k (1< k<= n) is the light k-1.At time of 0, some of them turn on, and others turn off.
Change the state of light i (if it's on, turn off it; if it is not on, turn on it) at t+1 second (t >= 0), if the left of light i is on !!! Given the initiation state, please find all lights’ state after M second. (2<= n <= 100, 1<= M<= 10^8)

 

Input

The input contains one or more data sets. The first line of each data set is an integer m indicate the time, the second line will be a string T, only contains '0' and '1' , and its length n will not exceed 100. It means all lights in the circle from 1 to n.
If the ith character of T is '1', it means the light i is on, otherwise the light is off.

 

Output

For each data set, output all lights' state at m seconds in one line. It only contains character '0' and '1.

 

Sample Input

   
   
   
   

    
    
    
    1
0101111
10
100000001
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1111000
001000010
   
   
   
   

 

解题思路：问题是给一排灯的开关状态，如果左边的灯前一秒是亮着的，那么该灯下一秒就变换状态，否则不变。问m秒后的情况。显然，第i盏灯在某一秒的状态ai，在下一秒时状态应该为（ai+ai-1）% 2，其实就是与左边进行异或运算。那么这道题目就很好利用状态矩阵来算了,构造好矩阵在m次方即可。。

这道题目在hdu上可以A,但在nyoj上就超时了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef long long LL;
const int mod = 2;
struct Matrix
{
	int m[100][100];
};
Matrix I,base;
int v[100],n;
char str[100];

void init()
{
	for(int i = 0; i < 100; i++)
		I.m[i][i] = 1;
}

Matrix mul(Matrix a,Matrix b)
{
	Matrix c;
	for(int i = 0; i < n; i++)
		for(int j = 0; j < n; j++)
		{
			c.m[i][j] = 0;
			for(int k = 0; k < n; k++)
			{
				c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]) % mod;
			}
		}
	return c;
}

Matrix power(Matrix A,int k)
{
	Matrix ans = I;
	while(k)
	{
		if(k & 1) ans = mul(ans,A);
		A = mul(A,A);
		k >>= 1;
	}
	return ans;
}

void build_matrix()	//构建n*n的矩阵
{
	for(int i = 0; i < n; i++)
		for(int j = 0; j < n; j++)
		{
			if(i == 0)
			{
				if(j == 0 || j == n-1)
					base.m[i][j] = 1;
				else base.m[i][j] = 0;
			}
			else
			{
				if(j == i-1 || j == i)
					base.m[i][j] = 1;
				else base.m[i][j] = 0;
			}
		}
}

int main()
{	
	init();
	LL m;
	while(scanf("%I64d",&m)!=EOF)
	{
		getchar();
		gets(str);
		n = strlen(str);
		build_matrix();
		for(int i = 0; i < n; i++)
			v[i] = str[i] - '0';
		Matrix A = power(base,m);
		int ans[100];
		for(int i = 0; i < n; i++)
		{
			ans[i] = 0;
			for(int k = 0; k < n; k++)
			{
				ans[i] = (ans[i] + A.m[i][k]*v[k]) % mod;
			}
		}
		for(int i = 0; i < n; i++)
			printf("%d",ans[i]);
		printf("\n");
	}
	return 0;
}