Kiki & Little Kiki 2
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1659 Accepted Submission(s): 837
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)
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.
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
Recommend
lcy
1 /* 2 题意：一个灯环，输出进行N此变化的结果，变化的规则是：如果左边是1，那么自己变成相反的 3 1变成0 0变成1 。第一个要以最后一个为参考。 4 5 矩阵的变化， 6 7 | 1 0 0 0 0 1| a1 | （a1+a6）%2 | 8 | 1 1 0 0 0 0| a2 | （a1+a2）%2 | 9 | 0 1 1 0 0 0| a3 === | （a2+a3）%2 | 10 | 0 0 1 1 0 0| a4 | （a3+a4）%2 | 11 | 0 0 0 1 1 0| a5 | （a4+a5）%2 | 12 | 0 0 0 0 1 1| a6 | （a5+a6）%2 | 13 14 利用矩阵相乘就可以快速到求取出来. 15 优化1.由于左边是稀疏矩阵，那么在矩阵相乘到时候，可以优化。代码中有！！~ 16 优化2.用位操作。怎么用位操作实现？ 17 优化3.在快速幂的时候，顺序的变化。因为稀疏矩阵的存在，所以改变一下顺序 18 就会有提高。代码中有... 19 */ 20 21 #include<iostream> 22 #include<cstdio> 23 #include<cstring> 24 #include<cstdlib> 25 using namespace std; 26 27 28 char a[103]; 29 struct node 30 { 31 int mat[103][103]; 32 }M_tom,M_hxl; 33 34 void make_first(node *cur,int len) 35 { 36 int i,j; 37 for(i=1;i<=len;i++) 38 for(j=1;j<=len;j++) 39 if(i==j) cur->mat[i][j]=1; 40 else cur->mat[i][j]=0; 41 } 42 43 void cheng2(node cur,char a[],int len) 44 { 45 node ww; 46 int i,j,k; 47 memset(ww.mat,0,sizeof(ww.mat)); 48 for(i=1;i<=len;i++) 49 for(k=1;k<=len;k++) 50 if(cur.mat[i][k]) 51 { 52 for(j=1;j<=1;j++) 53 { 54 ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&(a[k]-'0')))&1; 55 } 56 } 57 for(i=1;i<=len;i++) 58 printf("%d",ww.mat[i][1]); 59 printf(" "); 60 } 61 62 struct node cheng(node cur,node now,int len) 63 { 64 node ww; 65 int i,j,k; 66 memset(ww.mat,0,sizeof(ww.mat)); 67 for(i=1;i<=len;i++) 68 for(k=1;k<=len;k++) 69 if(cur.mat[i][k])//对稀疏矩阵的优化 70 { 71 for(j=1;j<=len;j++) 72 { 73 if(now.mat[k][j])//同理 74 { 75 ww.mat[i][j]=(ww.mat[i][j]^(cur.mat[i][k]&now.mat[k][j]))&1; 76 } 77 } 78 } 79 return ww; 80 } 81 82 void make_init(int len) 83 { 84 85 for(int i=1;i<=len;i++) 86 M_hxl.mat[1][i]=0; 87 88 M_hxl.mat[1][1]=1; 89 M_hxl.mat[1][len]=1; 90 for(int i=2;i<=len;i++) 91 for(int j=1;j<=len;j++) 92 if(i==j || i-1==j) 93 M_hxl.mat[i][j]=1; 94 else M_hxl.mat[i][j]=0; 95 } 96 97 98 void power_sum2(int n,int len,char a[]) 99 { 100 make_first(&M_tom,len); 101 while(n) 102 { 103 if(n&1) 104 { 105 M_tom=cheng(M_hxl,M_tom,len);//这也是优化。顺序改变结果就不同了 106 } 107 n=n>>1; 108 M_hxl=cheng(M_hxl,M_hxl,len); 109 } 110 // cs(len); 111 cheng2(M_tom,a,len); 112 } 113 114 int main() 115 { 116 int n,len; 117 while(scanf("%d",&n)>0) 118 { 119 scanf("%s",a+1); 120 len=strlen(a+1); 121 make_init(len); 122 power_sum2(n,len,a); 123 } 124 return 0; 125 }