hdu5863 cjj's string game
2016-08-21 11:28
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矩阵快速幂
#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
int N,M,K;
struct Matrix{
int a[25][25];
int n;
Matrix(int _n=0){
memset(a,0,sizeof(a));
n = _n;
}
Matrix operator *(const Matrix &T) const{
Matrix ans = Matrix(n);
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j)
for(int k = 0; k < n; ++k)
ans.a[i][j] = (ans.a[i][j] + 1ll*a[i][k]*T.a[k][j]%MOD) % MOD;
return ans;
}
};
Matrix operator ^(Matrix x,int y){
Matrix ans = Matrix(x.n);
for(int i = 0; i < x.n; ++i) ans.a[i][i] = 1;
while(y){
if(y&1) ans=ans*x;
y >>= 1; x = x*x;
}
return ans;
}
int main(){
int T; scanf("%d",&T);
while(T--) {
scanf("%d %d %d",&N,&M,&K);
Matrix A = Matrix(M*2+1);
A.a[0][0] = 1;
Matrix P = Matrix(M*2+1);
for(int i = 0; i < M; ++i) P.a[i][0] = K*K-K;
for(int i = 1; i < M; ++i) P.a[i-1][i] = K;
P.a[M-1][M*2] = K; P.a[M*2-1][M*2] = K;
for(int i = M; i < M*2+1; ++i) P.a[i][M] = K*K-K;
for(int i = M+1; i < M*2; ++i) P.a[i-1][i] = K;
P = P^N;
P = A*P;
int ans = 0;
for(int i = M; i < M*2+1; ++i) ans = (ans + P.a[0][i]) % MOD;
printf("%d\n",ans);
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
int N,M,K;
struct Matrix{
int a[25][25];
int n;
Matrix(int _n=0){
memset(a,0,sizeof(a));
n = _n;
}
Matrix operator *(const Matrix &T) const{
Matrix ans = Matrix(n);
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j)
for(int k = 0; k < n; ++k)
ans.a[i][j] = (ans.a[i][j] + 1ll*a[i][k]*T.a[k][j]%MOD) % MOD;
return ans;
}
};
Matrix operator ^(Matrix x,int y){
Matrix ans = Matrix(x.n);
for(int i = 0; i < x.n; ++i) ans.a[i][i] = 1;
while(y){
if(y&1) ans=ans*x;
y >>= 1; x = x*x;
}
return ans;
}
int main(){
int T; scanf("%d",&T);
while(T--) {
scanf("%d %d %d",&N,&M,&K);
Matrix A = Matrix(M*2+1);
A.a[0][0] = 1;
Matrix P = Matrix(M*2+1);
for(int i = 0; i < M; ++i) P.a[i][0] = K*K-K;
for(int i = 1; i < M; ++i) P.a[i-1][i] = K;
P.a[M-1][M*2] = K; P.a[M*2-1][M*2] = K;
for(int i = M; i < M*2+1; ++i) P.a[i][M] = K*K-K;
for(int i = M+1; i < M*2; ++i) P.a[i-1][i] = K;
P = P^N;
P = A*P;
int ans = 0;
for(int i = M; i < M*2+1; ++i) ans = (ans + P.a[0][i]) % MOD;
printf("%d\n",ans);
}
return 0;
}
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