题解以后再写
- T1：CF1117D Magic Gems
- T2：CF821E Okabe and El Psy Kongroo
- T3：CF954F Runner's Problem
- T4：CF1182E Product Oriented Recurrence
- T5：CF1151F Sonya and Informatics
T1：CF1117D Magic Gems
title
code
#include
#include
#define mod 1000000007
#define ll long long
int m;
ll n;
struct Matrix {
ll c[105][105];
Matrix () {
memset( c, 0, sizeof( c ) );
}
Matrix operator * ( const Matrix &p ) const {
Matrix ans;
for( int i = 1;i <= m;i ++ )
for( int j = 1;j <= m;j ++ )
for( int k = 1;k <= m;k ++ )
ans.c[i][j] = ( ans.c[i][j] + c[i][k] * p.c[k][j] % mod ) % mod;
return ans;
}
}A, result;
Matrix qkpow( Matrix a, ll b ) {
Matrix ans;
for( int i = 1;i <= m;i ++ ) ans.c[i][i] = 1;
while( b ) {
if( b & 1 ) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
int main() {
scanf( "%lld %d", &n, &m );
for( int i = 2;i <= m;i ++ )
A.c[i][i - 1] = 1;
A.c[1][m] = A.c[m][m] = 1;
result.c[1][m] = 1;
result = result * qkpow( A, n );
printf( "%lld", result.c[1][m] );
}
T2：CF821E Okabe and El Psy Kongroo
title
code
#include
#include
#include
using namespace std;
#define ll long long
#define mod 1000000007
struct Matrix {
int n, m;
ll c[16][16];
void clear() {
n = m = 0;
memset( c, 0, sizeof( c ) );
}
Matrix operator * ( const Matrix &p ) const {
Matrix ans;
ans.clear();
ans.n = n, ans.m = p.m;
for( int i = 0;i <= n;i ++ )
for( int k = 0;k <= m;k ++ )
for( int j = 0;j <= p.m;j ++ )
ans.c[i][j] = ( ans.c[i][j] + c[i][k] * p.c[k][j] % mod ) % mod;
return ans;
}
}v, result;
Matrix qkpow( Matrix a, ll b ) {
Matrix ans;
ans.clear();
ans.n = ans.m = a.n;
for( int i = 0;i <= ans.n;i ++ )
ans.c[i][i] = 1;
while( b ) {
if( b & 1 ) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
int n;
ll k;
int main() {
result.clear();
result.n = 15, result.m = 0;
result.c[0][0] = 1;
scanf( "%d %lld", &n, &k );
for( int i = 1;i <= n;i ++ ) {
ll a, b; int c;
scanf( "%lld %lld %d", &a, &b, &c );
b = min( b, k );
v.clear();
v.n = v.m = 15;
for( int j = 0;j <= c;j ++ ) {
v.c[j][j] = 1;
if( j ) v.c[j][j - 1] = 1;
if( j < c ) v.c[j][j + 1] = 1;
}
v = qkpow( v, b - a );
result = result * v;
}
printf( "%lld", result.c[0][0] );
return 0;
}
T3：CF954F Runner’s Problem
title
code
#include
#include
#include
using namespace std;
#define mod 1000000007
#define ll long long
struct Matrix {
ll c[3][3];
void clear () {
memset( c, 0, sizeof( c ) );
}
void zero () {
c[0][0] = c[1][1] = c[2][2] = c[0][1] = c[1][0] = c[1][2] = c[2][1] = 1;
}
Matrix operator * ( const Matrix &p ) const {
Matrix ans;
ans.clear();
for( int i = 0;i < 3;i ++ )
for( int k = 0;k < 3;k ++ )
for( int j = 0;j < 3;j ++ )
ans.c[i][j] = ( ans.c[i][j] + c[i][k] * p.c[k][j] % mod ) % mod;
return ans;
}
};
struct node {
int a; ll l, r;
}a[10005];
ll b[20010], c[3][20010], f[3];
Matrix qkpow( Matrix x, ll y ) {
Matrix v;
v.clear();
v.c[0][0] = v.c[1][1] = v.c[2][2] = 1;
while( y ) {
if( y & 1 ) v = v * x;
x = x * x, y >>= 1;
}
return v;
}
int n, num;
ll m;
int main() {
scanf( "%d %lld", &n, &m );
for( int i = 1;i <= n;i ++ ) {
scanf( "%d %lld %lld", &a[i].a, &a[i].l, &a[i].r );
a[i].a --;
b[++ num] = a[i].l - 1;
b[++ num] = a[i].r;
}
b[++ num] = 1, b[++ num] = m;
sort( b + 1, b + num + 1 );
num = unique( b + 1, b + num + 1 ) - b - 1;
for( int i = 1;i <= n;i ++ ) {
int l = lower_bound( b + 1, b + num + 1, a[i].l ) - b;
int r = lower_bound( b + 1, b + num + 1, a[i].r ) - b;
c[a[i].a][l] ++, c[a[i].a][r + 1] --;
}
Matrix ans;
ans.clear();
ans.c[1][0] = 1;
for( int i = 2;i <= num;i ++ ) {
Matrix v;
v.clear(), v.zero();
for( int j = 0;j < 3;j ++ ) {
f[j] += c[j][i];
if( f[j] ) v.c[j][0] = v.c[j][1] = v.c[j][2] = 0;
}
v = qkpow( v, b[i] - b[i - 1] );
ans = v * ans;
}
printf( "%lld", ans.c[1][0] );
return 0;
}
T4：CF1182E Product Oriented Recurrence
title
code
#include
#include
#define ll long long
#define mod 1000000007
struct Matrix {
int n, m;
ll c[6][6];
void clear() {
n = m = 0;
memset( c, 0, sizeof( c ) );
}
Matrix operator * ( const Matrix &p ) const {
Matrix ans;
ans.clear();
ans.n = n, ans.m = p.m;
for( int i = 1;i <= n;i ++ )
for( int k = 1;k <= m;k ++ )
for( int j = 1;j <= p.m;j ++ )
ans.c[i][j] = ( ans.c[i][j] + c[i][k] * p.c[k][j] % ( mod - 1 ) ) % ( mod - 1 );
return ans;
}
}v, p, w;
Matrix qkpow( Matrix x, ll y ) {
Matrix ans;
ans.clear();
ans.n = x.n, ans.m = x.m;
for( int i = 1;i <= ans.n;i ++ )
ans.c[i][i] = 1;
while( y ) {
if( y & 1 ) ans = ans * x;
x = x * x, y >>= 1;
}
return ans;
}
ll qkpow( ll x, ll y ) {
ll ans = 1;
while( y ) {
if( y & 1 ) ans = ans * x % mod;
x = x * x % mod, y >>= 1;
}
return ans;
}
ll n, f1, f2, f3, c, result = 1;
int main() {
scanf( "%lld %lld %lld %lld %lld", &n, &f1, &f2, &f3, &c );
w.clear(), v.clear();
w.n = 5, w.m = 1, w.c[5][1] = 2;
v.n = v.m = 5;
v.c[1][1] = v.c[1][2] = v.c[1][3] = v.c[1][4] = v.c[1][5]
= v.c[2][1] = v.c[3][2] = v.c[4][4] = v.c[4][5] = v.c[5][5] = 1;
v = qkpow( v, n - 3 );
w = v * w;
result = result * qkpow( c, w.c[1][1] ) % mod;
v.clear();
v.n = v.m = 3;
v.c[1][1] = v.c[1][2] = v.c[1][3] = v.c[2][1] = v.c[3][2] = 1;
v = qkpow( v, n - 3 );
p.clear();
p.n = 3, p.m = 1;
p.c[3][1] = 1;
p = v * p;
result = result * qkpow( f1, p.c[1][1] ) % mod;
p.clear();
p.n = 3, p.m = 1;
p.c[2][1] = 1;
p = v * p;
result = result * qkpow( f2, p.c[1][1] ) % mod;
p.clear();
p.n = 3, p.m = 1;
p.c[1][1] = 1;
p = v * p;
result = result * qkpow( f3, p.c[1][1] ) % mod;
printf( "%lld", result );
return 0;
}
T5：CF1151F Sonya and Informatics
title
code
#include
#include
#define mod 1000000007
#define ll long long
struct Matrix {
ll c[101][101];
void clear () {
memset( c, 0, sizeof( c ) );
}
Matrix operator * ( const Matrix &p ) const {
Matrix ans;
ans.clear();
for( int i = 0;i <= 100;i ++ )
for( int k = 0;k <= 100;k ++ )
for( int j = 0;j <= 100;j ++ )
ans.c[i][j] = ( ans.c[i][j] + c[i][k] * p.c[k][j] % mod ) % mod;
return ans;
}
}v, p;
Matrix qkpow( Matrix x, ll y ) {
Matrix ans;
ans.clear();
for( int i = 0;i <= 100;i ++ )
ans.c[i][i] = 1;
while( y ) {
if( y & 1 ) ans = ans * x;
x = x * x, y >>= 1;
}
return ans;
}
ll qkpow( ll x, ll y ) {
ll ans = 1;
while( y ) {
if( y & 1 ) ans = ans * x % mod;
x = x * x % mod, y >>= 1;
}
return ans;
}
ll n, k, cnt;
int a[105];
int main() {
scanf( "%lld %lld", &n, &k );
for( int i = 1;i <= n;i ++ )
scanf( "%d", &a[i] ), cnt += ( a[i] == 0 );
int tot = 0;
for( int i = 1;i <= cnt;i ++ )
tot += ( a[i] == 0 );
p.clear(), v.clear();
p.c[0][tot] = 1;
for( ll i = 0;i <= cnt;i ++ ) {
if( i ) v.c[i - 1][i] = ( cnt - i + 1 ) * ( cnt - i + 1 ) % mod;
v.c[i][i] = ( i * ( cnt - i ) % mod + ( cnt - i ) * ( n - cnt - cnt + i ) % mod ) % mod;
v.c[i][i] = ( v.c[i][i] + cnt * ( cnt - 1 ) / 2 % mod ) % mod;
v.c[i][i] = ( v.c[i][i] + ( n - cnt ) * ( n - cnt - 1 ) / 2 % mod ) % mod;
if( i < cnt ) v.c[i + 1][i] = ( i + 1 ) * ( n - cnt - cnt + i + 1 ) % mod;
}
v = qkpow( v, k );
p = p * v;
ll ans = p.c[0][cnt], temp = 0;
for( int i = 0;i <= cnt;i ++ )
temp = ( temp + p.c[0][i] ) % mod;
printf( "%lld", ans * qkpow( temp, mod - 2 ) % mod );
return 0;
}
