Kiki & Little Kiki 2(找规律+矩阵快速幂的应用)
2015-08-06 00:49
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Link:http://acm.hdu.edu.cn/showproblem.php?pid=2276
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2209 Accepted Submission(s): 1134
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
Sample Output
Source
HDU 8th Programming Contest Site(1)
算法思想:
由题意,可得如下规律:
00 -----> ?0
10 -----> ?1
01 -----> ?1
11 -----> ?0
也就是说对于当前第i-1位、第i位字符,下一秒第i位字符的变化规律是:00->0 , 10->1 , 01->1 , 11->0 观察可发现下一秒第i位字符的结果是当前第i-1位、第i位字符进行异或运算后的结果。因此可以将原来长度为N的01字符串看成一个1xN的矩阵A,构造然后利用上面得出的元素之间的递推规律一个NxN的01方阵。比如N=7时,可构造矩阵B如下:
1 1 0 0 0 0 0
0 1 1 0 0 0 0
0 0 1 1 0 0 0
0 0 0 1 1 0 0
0 0 0 0 1 1 0
0 0 0 0 0 1 1
1 0 0 0 0 0 1
该矩阵构造原理如下:矩阵B中第i列中元素B[i][i]为1,其关联的是原字符串中的第i个字符,而第i列中另一个元素1关联的是原字符串中的第i-1个字符(题目中有说,第一个字符的前一个字符是第N个字符,所以第一列中另外一个元素1位于第N行,其他列中的两个1都是相邻行的)。
而异或运算如何通过矩阵乘法来实现呢?其实矩阵相乘时,就是把原来的数据与现在的数据相加了,我们只要将所得的数据模2即可。为什么这个可以替代异或呢?其实异或本来就是朴素的加法(二进制的),只不过1+1=10,我们取得是最低位罢了,所以可以直接模2就行了。
AC code:
Kiki & Little Kiki 2
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2209 Accepted Submission(s): 1134
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
Source
HDU 8th Programming Contest Site(1)
算法思想:
由题意,可得如下规律:
00 -----> ?0
10 -----> ?1
01 -----> ?1
11 -----> ?0
也就是说对于当前第i-1位、第i位字符,下一秒第i位字符的变化规律是:00->0 , 10->1 , 01->1 , 11->0 观察可发现下一秒第i位字符的结果是当前第i-1位、第i位字符进行异或运算后的结果。因此可以将原来长度为N的01字符串看成一个1xN的矩阵A,构造然后利用上面得出的元素之间的递推规律一个NxN的01方阵。比如N=7时,可构造矩阵B如下:
1 1 0 0 0 0 0
0 1 1 0 0 0 0
0 0 1 1 0 0 0
0 0 0 1 1 0 0
0 0 0 0 1 1 0
0 0 0 0 0 1 1
1 0 0 0 0 0 1
该矩阵构造原理如下:矩阵B中第i列中元素B[i][i]为1,其关联的是原字符串中的第i个字符,而第i列中另一个元素1关联的是原字符串中的第i-1个字符(题目中有说,第一个字符的前一个字符是第N个字符,所以第一列中另外一个元素1位于第N行,其他列中的两个1都是相邻行的)。
而异或运算如何通过矩阵乘法来实现呢?其实矩阵相乘时,就是把原来的数据与现在的数据相加了,我们只要将所得的数据模2即可。为什么这个可以替代异或呢?其实异或本来就是朴素的加法(二进制的),只不过1+1=10,我们取得是最低位罢了,所以可以直接模2就行了。
AC code:
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<queue> #include<map> #define LL long long #define MAXN 1000010 using namespace std; const int INF=0x3f3f3f3f; char s[111]; //----以下为矩阵快速幂模板-----// //const int mod=1000;//模3,故这里改为3即可 int mod=2; const int NUM=101;//定义矩阵能表示的最大维数 int N;//N表示矩阵的维数,以下的矩阵加法、乘法、快速幂都是按N维矩阵运算的 struct Mat{//矩阵的类 int a[NUM][NUM]; Mat(){memset(a,0,sizeof(a));} void init()//将其初始化为单位矩阵 { memset(a,0,sizeof(a)); for(int i=0;i<NUM;i++) { a[i][i]=1; } } }; Mat A,B,mi,ans; Mat add(Mat a,Mat b)//(a+b)%mod 矩阵加法 { Mat ans; for(int i=0;i<N;i++) { for(int j=0;j<N;j++) { ans.a[i][j]=(a.a[i][j]%mod)+(b.a[i][j]%mod); ans.a[i][j]%=mod; } } return ans; } Mat mul(Mat a,Mat b) //(a*b)%mod 矩阵乘法 { Mat ans; for(int i=1;i<=N;i++) { for(int j=1;j<=N;j++) { ans.a[i][j]=0; for(int k=1;k<=N;k++) { ans.a[i][j]=(ans.a[i][j]+a.a[i][k]*b.a[k][j])%2; } //ans.a[i][j]%=mod; } } return ans; } Mat power(Mat a,int num)//(a^n)%mod 矩阵快速幂 { Mat ans; ans.init(); while(num) { if(num&1) { ans=mul(ans,a); } num>>=1; a=mul(a,a); } return ans; } Mat pow_sum(Mat a,int num)//(a+a^2+a^3....+a^n)%mod 矩阵的幂和 { int m; Mat ans,pre; if(num==1) return a; m=num/2; pre=pow_sum(a,m); ans=add(pre,mul(pre,power(a,m))); if(num&1) ans=add(ans,power(a,num)); return ans; } void output(Mat a)//输出矩阵 { for(int i=1;i<=N;i++) { for(int j=1;j<=N;j++) { printf("%d%c",a.a[i][j],j==N-1?'\n':' '); } } } //----以上为矩阵快速幂模板-----// int main() { //freopen("D:\in.txt","r",stdin); int n,m,i,j; while(scanf("%d",&m)!=EOF) { scanf("%s",s); n=strlen(s); N=n; memset(A.a,0,sizeof(A.a)); memset(B.a,0,sizeof(B.a)); for(i=1;i<=N;i++) { A.a[1][i]=s[i-1]-'0'; } for(i=1;i<=N-1;i++) { B.a[i][i]=B.a[i][i+1]=1; } B.a [1]=B.a =1; B=power(B,m); for(int i=1;i<=1;i++) { for(int j=1;j<=N;j++) { ans.a[i][j]=0; for(int k=1;k<=N;k++) { ans.a[i][j]=(ans.a[i][j]+A.a[i][k]*B.a[k][j])%2; } //ans.a[i][j]%=mod; } } for(i=1;i<=N;i++) printf("%d",ans.a[1][i]); printf("\n"); } return 0; }
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