test bottom tro little seconds sta struct 第一次 bold

Kiki & Little Kiki 2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2650 Accepted Submission(s): 1393

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） 一開始沒想到公式。。。看了一眼別人的公示後推開了矩陣。。腦子秀逗了總把乘號寫成加號 公式a[i]=(a[i]+a[i-1])%2;,特別的對於a[1]=(a[i]+a[N])%2 M最大10億顯然樸素法不可取，由於是看專題進來的所以直接想的就是矩陣= = 假設第零次的數組為原始01串(a,b,c,d) 則第一次 ((a+d)%2,(b+a)%2,(c+b)%2,(d+c)%2) 第二次 (((a+d)%2+(d+c)%2)%2,......) 不難構造出一個N*N的矩陣，第i列的第i和i-1個元素置為1其余元素置0即可。 然後計算出這個轉移矩陣的N次冪後再與原始行矩陣相乘得到答案。 此處還用到了同余定理 (A+B)%M=(A%M+B%M)%M; #include<bits/stdc++.h>
using namespace std;
int N,M;
struct Matrix
{
int a[105][105];
Matrix operator*(Matrix tmp){
Matrix ans;
memset(ans.a,0,sizeof(ans.a));
for(int i=1;i<=N;++i){
for(int k=1;k<=N;++k){
for(int j=1;j<=N;++j){
ans.a[i][j]+=a[i][k]*tmp.a[k][j];
ans.a[i][j]%=2;
}
}
}
return ans;
}
};
void show(Matrix a)
{int i,j,k;
for(i=1;i<=N;++i){
for(j=1;j<=N;++j){
cout<<a.a[i][j]<<" ";
}cout<<endl;
}cout<<endl;
}
Matrix qpow(Matrix A,int n)
{
Matrix ans;
memset(ans.a,0,sizeof(ans.a));
for(int i=0;i<=N;++i) ans.a[i][i]=1;
while(n){
if(n&1) ans=ans*A;
A=A*A;
n>>=1;
}
return ans;
}

void solve(string s)
{
Matrix A;
int i,j,k,u[105];
char ans[105];
for(i=0;i<s.size();++i) u[i+1]=s[i]-‘0‘;
memset(A.a,0,sizeof(A.a));
A.a[1][1]=A.a[N][1]=1;
for(i=2;i<=N;++i){
A.a[i-1][i]=A.a[i][i]=1;
}
A=qpow(A,M);

for(i=1;i<=N;++i){int d=0;
for(j=1;j<=N;++j){
d+=u[j]*A.a[j][i];
d%=2;
}
cout<<d%2;
}cout<<endl;
}
int main()
{
string s;
while(cin>>M>>s){
N=s.size();
solve(s);
}
return 0;
}

HDU 2276 矩陣快速冪