[JSC2021 A~D + F]
半小时打完了\(A~D\)，想要一发\(F\)冲进前\(100\),结果平衡树常数大\(T\)了。据说\(G\)是矩阵树定。
\(A\)
放代码吧。
A
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll x,y,z;
int main(){
x = read(),y = read(),z = read();
if(y * z % x != 0)
std::cout<<(ll)y * z / x;
else
std::cout<<(ll)y * z / x - 1<<std::endl;
}
\(B\)
按题意模拟
B
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
#define M 1000000
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll num[M];
ll n,m;
int main(){
n = read(),m = read();
for(int i = 1;i <= n;++i){
ll x = read();
if(num[x] == 0)
num[x] ++ ;
}
for(int i = 1;i <= m;++i){
ll x = read();
num[x] ++ ;
}
for(int i = 1;i <= M;++i)
if(num[i] == 1)
std::cout<<i<<" ";
}
\(C\)
考虑枚举这个最大的公约数，把这个公约数的倍数在小于\(m\)情况下求出第二大的,看是否大于\(n\)
C
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
#define M 1000000
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll num[M];
ll n,m;
bool s(int a){
for(int i = 2;i < sqrt(a);++i)
if(a % i == 0)
return false;
return true;
}
int main(){
n = read(),m = read();
for(int i = m;i >= 1;--i){
ll x = m / i * i - i;
ll y = m / i * i;
if(x >= n){
std::cout<<i;
return 0;
}
}
}
\(D\)
考虑第一位有\(p - 1\)种选择，以后每个位根据前面的和的膜，只有\(p-2\)种
D
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll n,p;
ll power(ll a,ll b){
ll ans = 1;
while(b){
if(b & 1)ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
int main(){
n = read();
p = read();
p -= 1;
std::cout<<(p * power(p - 1,n - 1)) % mod<<std::endl;
}
\(F\)
考虑改变一个数时，在另外一个序列里找到原数的贡献，和现在这个数的贡献。
并在这个序列中删掉原数，加入新数。
用平衡树操作，只要查询前缀，前缀和，还有删除插入操作
放一下考场代码，被卡常了，找时间再改吧。
F
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
#define M 200005
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll n,m,q;
ll a[M],b[M],suma,sumb;
ll ans = 0;
struct P{
#define A 2000010
ll ch[A][2],val[A],cv[A],siz[A],cnt,sum[A];
#define l(x) ch[x][0]
#define r(x) ch[x][1]
#define v(x) val[x]
#define c(x) cv[x]
#define s(x) siz[x]
#define sa(x) sum[x]
void up(ll x){s(x) = 1 + s(l(x)) + s(r(x)),sa(x) = v(x) + sa(l(x)) + sa(r(x));}
ll randoom(){return rand() << 15 | rand();}
ll newcode(ll x){s(++cnt) = 1,sa(cnt) = v(cnt) = x,c(cnt) = randoom();return cnt;}
void split(ll now,ll k,ll &x,ll &y){
if(!now){x = y = 0;return;}
if(v(now) <= k) x = now,split(r(now),k,r(now),y);
else
y = now,split(l(now),k,x,l(now));
up(now);
}
ll merge(ll x,ll y){
if(!x || !y)return x + y;
if(c(x) < c(y)){
r(x) = merge(r(x),y);
up(x);return x;
}
else{
l(y) = merge(x,l(y));
up(y);return y;
}
}
ll root,x,y,z,cn;
void insert(ll a){
cn ++ ;
split(root,a,x,y);
root = merge(merge(x,newcode(a)),y);
}
void del(ll a){
cn -- ;
split(root,a,x,z);
split(x,a - 1,x,y);
y = merge(l(y),r(y));
root = merge(x,merge(y,z));
}
ll find(ll a){
split(root,a - 1,x,y);
ll ans = s(x) + 1;
root = merge(x,y);
return ans;
}
ll kth(ll now,ll k){
if(k <= s(l(now)))return kth(l(now),k);
else
if(k == s(l(now)) + 1)return now;
else
return kth(r(now),k - s(l(now)) - 1);
}
ll pre(ll a){
split(root,a,x,y);
ll ans = s(x);
merge(x,y);
return ans;
}
ll nex(ll a){
split(root,a,x,y);
ll ans = v(kth(y,1));
merge(x,y);
return ans;
}
ll prev(ll a){
split(root,a,x,y);
ll ans = sa(kth(x,s(x)));
merge(x,y);
return ans;
}
}Q,P;//A B
ll tob(int x){
ll l = sumb - P.prev(x);
// std::cout<<l<<std::endl;
ll k = P.pre(x) * x;
// std::cout<<k<<std::endl;
return l + k;
}//在b里找贡献。
ll toa(int x){
ll l = suma - Q.prev(x);
ll k = Q.pre(x) * x;
return l + k;
}//在b里找贡献。
int main(){
n = read(),m = read(),q = read();
for(int i = 1;i <= n;++i)
Q.insert(0);
for(int i = 1;i <= m;++i)
P.insert(0);
while(q -- ){
ll t = read(),x = read(),y = read();
if(t == 1){
ans += tob(y) - tob(a[x]);
suma += y - a[x];
Q.del(a[x]),Q.insert(y);
a[x] = y;
}
if(t == 2){
ans += toa(y) - toa(b[x]);
sumb += y - b[x];
P.del(b[x]),P.insert(y);
b[x] = y;
}
std::cout<<ans<<std::endl;
}
}
\(rank 800\)果真还是逊啊。
