Codeforces Round #323 (Div. 2) D 582B Once Again...（快速幂）

A[i][j]表示在循环节下标i开头j结尾的最长不减子序列，这个序列的长度为p,另外一个长度为q的序列对应的矩阵为B[i][j]，

将两序列合并，新的序列对应矩阵C[i][j] = max(A[i][k]+B[k][j])。非法的情况标记为-INF，用倍增加速。

#include<bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 101;
int n;
typedef int MType;
struct Matrix
{
MType dat[maxn][maxn];
MType *operator [](int x){ return dat[x]; }
Matrix operator | (Matrix& B) {
Matrix re;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
re[i][j] = -INF;
for(int k = 0; k < n; k++){
re[i][j] = max(re[i][j],dat[i][k]+B[k][j]);
}
}
}
return re;
}
Matrix operator ^ (int q){
Matrix Re, A = *this;
memset(Re.dat,0,sizeof(Re.dat));
while(q){
if(q&1){
Re = Re | A;
}
A = A | A;
q >>= 1;
}
return Re;
}
};

int a[maxn];

//#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
int T; scanf("%d%d",&n,&T);
for(int i = 0; i < n; i++){
scanf("%d",a+i);
}

Matrix A;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(a[i]>a[j] ) A[i][j] = -INF;
else {
A[i][j] = 1;
for(int k = 0; k < j; k++){
if(a[k] <= a[j])
A[i][j] = max(A[i][j],A[i][k]+1);
}
}
}
}
A = A^T;
int ans = 0;
for(int i = 0; i < n;i++){
for(int j = 0; j < n; j++){
ans = max(ans,A[i][j]);
}
}
printf("%d\n",ans);
return 0;
}