hdu2276(矩阵的快速幂)

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1711    Accepted Submission(s): 874

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

1010111110100000001

 

Sample Output

1111000001000010

 

Source

HDU 8th Programming Contest Site（1）

 

Recommend

lcy

        本题所有的灯构成一个环，某时刻如果某盏灯的左边为on，则改变该灯的状态，问一段时间过后所有灯的状态。

        本题题意不难理解，时间过大，不能直接模拟。经过观察发现本题可以用矩阵的快速幂。分析见代码。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN=100+10;char str[MAXN];struct matrix{int len;int a[MAXN][MAXN];}Origin,res;/*********************************origin初始化为   1  0  0 …… 0  0  1   1  1  0 …… 0  0  0   0  1  1 …… 0  0  0   0  0  1 …… 0  0  0   ………………………   0  0  0 …… 1  1  0   0  0  0 …… 0  1  1res初始化为str的转置形式，其他位置填0*********************************/void init(int len){int i,j;memset(Origin.a,0,sizeof(Origin.a));for(i=0;i<len;i++){for(j=0;j<len;j++){if(i!=0){if(j==i||j==i-1)Origin.a[i][j]=1;else Origin.a[i][j]=0;}else {if(j==0||j==len-1)Origin.a[i][j]=1;else Origin.a[i][j]=0;}}}Origin.len=len;memset(res.a,0,sizeof(res.a));res.len=len;for(i=0;i<len;i++)res.a[i][0]=str[i]-'0';}void print(int len){int i,j;printf("*******Origin*******\n");for(i=0;i<len;i++){for(j=0;j<len;j++)printf("%d ",Origin.a[i][j]);printf("\n");}printf("*********res****\n");for(i=0;i<len;i++){for(j=0;j<len;j++)printf("%d ",res.a[i][j]);printf("\n");}}matrix MatrixMultiple(matrix &ori,matrix &ans){matrix ret;ret.len=ori.len;int i,j,k;memset(ret.a,0,sizeof(ret.a));for(i=0;i<ori.len;i++){for(j=0;j<ori.len;j++){for(k=0;k<ans.len;k++){ret.a[i][j]+=(ori.a[i][k]*ans.a[k][j]);}ret.a[i][j]%=2;}}return ret;}void MatrixPow(int t){while(t){if(t&1)res=MatrixMultiple(Origin,res);Origin=MatrixMultiple(Origin,Origin);t=t>>1;}}int main(){int t,len,i;while(~scanf("%d",&t)){scanf("%s",str);len=strlen(str);//printf("len=%d  str=%s\n",len,str);init(len);//print(len);MatrixPow(t);for(i=0;i<len;i++)printf("%d",res.a[i][0]);printf("\n");/*************************************直接模拟超时代吗k=0;//printf("t=%d  str=%s\n",t,str[k]);for(i=1;i<=t;i++){for(j=0;j<len;j++){//printf("***\n");if((j>0&&str[k][j-1]=='1')||(j==0&&str[k][len-1]=='1'))str[k^1][j]=(str[k][j]=='1'?'0':'1');else str[k^1][j]=str[k][j];}str[k^1][len]=0;k=k^1;//printf("t=%d  str=%s\n",t,str[k]);}*****************************************///printf("%s\n",str[k]);}system("pause");return 0;}