10.22 模拟赛
10.22 模拟赛
T1 染色
考虑每个连通块删成一棵树就好了。
mmp场上就我路径压缩写炸。。。。
f[1<<21],n,m,t;F(x){t=f[x];return x^t?f[x]=!t?x:F(t):x;}main(s){for(scanf("%*d%d",&s);~scanf("%d%d",&n,&m);f[F(m)]=F(n))s-=F(n)!=F(m);exit(!printf("%d",s));}
T2 乘方
首先，二分答案，变成一个判定问题。
如果只有一个集合，比一个数字小的数字个数就是 $ \sqrt[n]{m} $
然后如果是一些集合交，可以直接把这些集合取 $ lcm $ ，可以证明是对的。
然后可以 $ 2^k $ 枚举子集容斥
当然，这个容斥看起来就可以dp
然后dp就对了，设 $ dp[i][j] $ 表示前 i 个数，LCM为一个 j 的子集的容斥系数
记得开根号二分。。不然T惨
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define int long long
#define MAXN 62
typedef long long ll;
int fuck[MAXN][MAXN];
inline int gcd( int a , int b ) { return b ? gcd( b , a % b ) : a; }
inline int lcm( int x , int y ) {
if (fuck[x][y]) return fuck[x][y];
return fuck[x][y] = fuck[y][x] = x * y / gcd( x , y );
}
int m , k;
int A[MAXN];
int dp[MAXN][MAXN];
inline ll calc(ll n, int k)
{
ll val = 1;
for (int i = 0; i < k; i++)
{
if ((double)val * n >= 100000000000000000LL)
return 100000000000000000LL;
val *= n;
}
return val;
}
inline ll sqrt(ll n, int k)
{
ll res = std::pow(n, 1.0 / k);
while (calc(res, k) < n)
res++;
while (calc(res, k) > n)
res--;
return res - 1;
}
int qpow(int x,int y)
{
int res=1;
while(y)
{
if(y&1)
{
if(res*x<=0||res*x>1e17)return -1;
res*=x;
}
y>>=1;
if(!y)break;
if(x*x<=0||x*x>1e17)return -1;
x*=x;
}
return res;
}
int lim[MAXN];
int get(int x,int v)
{
int l=1,r=lim[x],res=1;
if(x>60)return 1;
while(l<=r)
{
int mid=(l+r)>>1;
int re=qpow(mid,x);
if(re<=v&&(re!=-1))res=mid,l=mid+1;
else r=mid-1;
}
return res;
}
inline bool chk( int x ) {
long long res = 0;
for( int n = 1 ; n < MAXN ; ++ n )
res += dp[k][n] * get( n , x ) - dp[k][n];
// cout << res << endl;
return res >= m;
}
signed main() {
lim[2]=1e9;lim[3]=1e6;lim[4]=sqrt(lim[2])+15;lim[5]=4500;lim[6]=2000;
for(int i=7;i<=60;++i)
{
for(int j=1;;++j)
{
int now=qpow(j,i);
if(now==-1)break;
lim[i]=j;
}
}
int t; cin >> t;
while( t-- ) {
scanf("%lld%lld",&m,&k), m--;
int flag = 0;
for( int i = 1 ; i <= k ; ++ i ) { scanf("%lld",&A[i]); if( A[i] == 1 ) flag = 1; }
if( flag ) { cout << m + 1 << endl; continue; }
memset( dp , 0 , sizeof dp );
dp[1][A[1]] = 1;
for( int i = 1 ; i < k ; ++ i ) {
dp[i + 1][A[i + 1]] = 1;
for( int n = 1 ; n < MAXN ; ++ n ) if (dp[i][n]) {
dp[i + 1][n] += dp[i][n];
if( lcm(n, A[i + 1]) < MAXN )
dp[i + 1][lcm(n, A[i + 1])] -= dp[i][n];
}
}
// for (int i = 2; i <= 50; i++)
// cout << dp[k][i][0] - dp[k][i][1] << ' ';
// cout << endl;
// chk(64);
ll l = 1 , r = 100000000000000006LL, ans = -1;
while( l <= r ) {
int mid = l + r >> 1;
if( chk( mid ) ) ans = mid, r = mid - 1;
else l = mid + 1;
}
cout << ans << endl;
}
}
T3 位运算
and 可以用 这里 的 F 的做法来做，复杂度 $ O(n + v) $ 当然也可以贪心 $ O(nlogv) $
xor 可以用trie直接做
or 不会。貌似只能FWT
所以就粘板子 FWT 跑就行了，直接求位运算卷积就做完了。
当 FWT 板子放在这里把
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
#include<set>
using namespace std;
//#define int long long
typedef long long ll;
#define MAXN 8388610
#define MAXV 8388610
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define inf 0x3f3f3f3f
#define cmx( a , b ) a = max( a , b )
#define cmn( a , b ) a = min( a , b )
#define upd( a , b ) ( a = ( a + b ) % P )
#define swap( a , b ) a = a ^ b , b = a ^ b , a = a ^ b
#define P 1000000007
int n , q;
int A[MAXN];
namespace fwt {
inline void FWT1(ll a[], int len) {
for (int mid = 2; mid <= len; mid <<= 1)
for (int i = 0; i < len; i += mid)
for (int j = i; j < i + (mid >> 1); j++)
a[j + (mid >> 1)] += a[j];
}
inline void IFWT1(ll a[], int len) {
for (int mid = 2; mid <= len; mid <<= 1)
for (int i = 0; i < len; i += mid)
for (int j = i; j < i + (mid >> 1); j++)
a[j + (mid >> 1)] -= a[j];
}
inline void FWT2(ll a[], int len) {
for (int mid = 2; mid <= len; mid <<= 1)
for (int i = 0; i < len; i += mid)
for (int j = i; j < i + (mid >> 1); j++)
a[j] += a[j + (mid >> 1)];
}
inline void IFWT2(ll a[], int len) {
for (int mid = 2; mid <= len; mid <<= 1)
for (int i = 0; i < len; i += mid)
for (int j = i; j < i + (mid >> 1); j++)
a[j] -= a[j + (mid >> 1)];
}
inline void FWT3(ll a[], int len) {
for (int mid = 2; mid <= len; mid <<= 1)
for (int i = 0; i < len; i += mid)
for (int j = i; j < i + (mid >> 1); j++) {
ll x = a[j], y = a[j + (mid >> 1)];
a[j] = x + y, a[j + (mid >> 1)] = x - y;
}
}
inline void IFWT3(ll a[], int len) {
for (int mid = 2; mid <= len; mid <<= 1)
for (int i = 0; i < len; i += mid)
for (int j = i; j < i + (mid >> 1); j++) {
ll x = a[j], y = a[j + (mid >> 1)];
a[j] = (x + y) >> 1, a[j + (mid >> 1)] = (x - y) >> 1;
}
}
ll a[MAXN], b[MAXN]; int input[400010];
int main() {
int mx = 0, len = 1;
for (int i = 1, x; i <= n; i++)
a[A[i]]++, b[A[i]]++, mx = max(mx, A[i]), input[i] = A[i];
while (len <= mx) len <<= 1;
if (q == 1) {
FWT2(a, len);
for (int i = 0; i < len; i++) a[i] = a[i] * a[i];
IFWT2(a, len);
for (int i = len - 1; i >= 0; i--) if (a[i]) {
ll res = (a[i] - b[i]) >> 1;
if (res) return printf("%d %lld", i, res), 0;
}
} else if (q == 2) {
bool flag = true;
for (int i = 2; i <= n; i++)
if (input[i] != input[1]) {
flag = false; break;
}
if (flag) return printf("0 %lld", (ll)n * (n + 1) / 2), 0;
FWT3(a, len);
for (int i = 0; i < len; i++) a[i] = a[i] * a[i];
IFWT3(a, len);
for (int i = len - 1; i >= 0; i--) if (a[i])
return printf("%d %lld", i, a[i] >> 1), 0;
} else {
FWT1(a, len);
for (int i = 0; i < len; i++) a[i] = a[i] * a[i];
IFWT1(a, len);
for (int i = len - 1; i >= 0; i--) if (a[i]) {
ll res = (a[i] - b[i]) >> 1;
if (res) return printf("%d %lld", i, res), 0;
}
}
}
}
int main() {
cin >> n >> q;
int mx = 0;
for( int i = 1 ; i <= n ; ++ i ) scanf("%d",&A[i]) , mx = max(A[i] , mx);
fwt::main();
}
