HDU 2276 矩阵快速幂
2017-07-11 21:19
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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
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) 一开始没想到公式。。。看了一眼别人的公示后推开了矩阵。。脑子秀逗了总把乘号写成加号 公式a[i]=(a[i]+a[i-1])%2;,特别的对于a[1]=(a[i]+a
)%2 M最大10亿显然朴素法不可取,由于是看专题进来的所以直接想的就是矩阵= = 假设第零次的数组为原始01串(a,b,c,d) 则第一次 ((a+d)%2,(b+a)%2,(c+b)%2,(d+c)%2) 第二次 (((a+d)%2+(d+c)%2)%2,......) 不难构造出一个N*N的矩阵,第i列的第i和i-1个元素置为1其余元素置0即可。 然后计算出这个转移矩阵的N次幂后再与原始行矩阵相乘得到答案。 此处还用到了同余定理 (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
[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;
}
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