hdu2276 Kiki & Little Kiki 2

最新推荐文章于 2020-11-11 23:26:07 发布

原创最新推荐文章于 2020-11-11 23:26:07 发布 · 1.8k 阅读

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本文介绍了一个基于矩阵乘法的灯状态模拟算法，通过快速幂运算高效计算圆形排列灯经过特定秒数后的状态变化。

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 551 Accepted Submission(s): 317

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

0101111

100000001

Sample Output

1111000

001000010

Source

HDU 8th Programming Contest Site（ 1 ）

Recommend

lcy

依旧是矩阵乘法

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，灯ai 下一秒的状态就是 (ai-1+ai) mod 2 ，所以可以构造如下矩阵

假设有5 盏灯

1 0 0 0 1

1 1 0 0 0

0 1 1 0 0

0 0 1 1 0

0 0 0 1 1

之后直接用快速幂解决。

代码如下

#include <stdio.h> #include <string.h> int a[105][105]; int ans[105]; int tag[105]; char str[105]; int tag1[105][105]; int main() { int i,j,m,n,k; while(scanf("%d",&m)!=EOF) { scanf("%s",str); n=strlen(str); for (i=0;i<n;i++) { for (j=0;j<n;j++) { if (j==i || j==(i+n-1)%n) a[i][j]=1; else a[i][j]=0; } } for (i=0;i<n;i++) { ans[i]=str[i]-'0'; } while(m!=0) { if (m%2==1) { for (i=0;i<n;i++) { tag[i]=0; for (j=0;j<n;j++) { tag[i]=(tag[i]+a[i][j]*ans[j])%2; } } for (i=0;i<n;i++) { ans[i]=tag[i]; } } m=m/2; for (i=0;i<n;i++) { for (j=0;j<n;j++) { tag1[i][j]=0; for (k=0;k<n;k++) { tag1[i][j]=(tag1[i][j]+a[i][k]*a[k][j])%2; } } } for (i=0;i<n;i++) { for (j=0;j<n;j++) { a[i][j]=tag1[i][j]; } } } for (i=0;i<n;i++) { printf("%d",ans[i]); } printf("/n"); } return 0; }