1. 819B Mister B and PR Shifts
大意: 给定排列$p$, 定义排列$p$的特征值为$\sum |p_i-i|$, 可以循环右移任意位, 求最小特征值和对应移动次数.
右移过程中维护增加的个数和减少的个数即可.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head const int N = 1e6+;
int n, a[N];
int dl[N], dr[N]; int main() {
scanf("%d",&n);
REP(i,,n) scanf("%d",a+i);
ll ret = , ans = ;
int L = , R = ;
REP(i,,n) {
ret += abs(a[i]-i);
if (a[i]>i) {
++L;
--dl[a[i]-i];
++dr[a[i]-i];
}
else if (a[i]<=i) {
++R;
if (a[i]!=) {
--dl[n-i+a[i]];
++dr[n-i+a[i]];
}
}
}
ans = ret;
int pos = ;
REP(i,,n-) {
ret += R-L;
R += dr[i];
L += dl[i];
if (a[n-i+]!=) --R,++L;
ret -= abs(a[n-i+]-n-);
ret += abs(a[n-i+]-);
if (ret<ans) pos = i, ans = ret;
}
printf("%lld %d\n", ans, pos);
}
2. 819C Mister B and Beacons on Field
大意: 给定两个平面点$A(m,0),B(0,n)$
- 求$A$移向原点过程中, 有多少个时刻, 存在一个点$C$使得ABC面积为$S$
- $A$在原点, $B$移向原点过程中, 有多少个时刻, 存在一个点$C$使得ABC面积为$S$
对于第一问, 假设$C$坐标为$(x,y)$, $A$返回时坐标为$(t,0)$
那么有$xn+ty=2S+tn, 0\le t\le m$
就等价于求$[0,m]$中有多少个$t$满足$gcd(n,t)|2S$
对于第二问, 假设$B$返回时坐标为$(0,t)$
那么有$xt=2S, 1\le t\le m$
等价于求$[1,n]$中有多少个$t$满足$t|2S$
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head const int N = 1e6+;
int gpf[N];
vector<pii> B,C,S;
ll ans,n,m,s; void solve(vector<pii> &A, int x) {
while (x!=) {
int p = gpf[x], cnt = ;
while (x%p==) x/=p,++cnt;
A.pb(pii(p,cnt));
}
}
void repr(vector<pii> &A) {
sort(A.begin(),A.end());
vector<pii> ret;
for (auto &t:A) {
if (ret.empty()||t.x!=ret.back().x) ret.pb(t);
else ret.back().y += t.y;
}
A = ret;
}
void init() {
int n1,n2,n3,m1,m2,m3,s1,s2,s3;
scanf("%d%d%d%d%d%d%d%d%d",&n1,&n2,&n3,&m1,&m2,&m3,&s1,&s2,&s3);
B.clear(),C.clear(),S.clear();
n = (ll)n1*n2*n3, m = (ll)m1*m2*m3, s = (ll)s1*s2*s3;
solve(B,n1),solve(B,n2),solve(B,n3),repr(B);
S.pb(pii(,));
solve(S,s1),solve(S,s2),solve(S,s3),repr(S);
}
void dfs(int d, ll num, int z) {
if (!num) return;
if (d==C.size()) return ans+=num*z,void();
dfs(d+,num,z);
REP(i,,C[d].y+) num/=C[d].x;
dfs(d+,num,-z);
}
void dfs2(int d, ll num) {
if (num>n) return;
if (d==S.size()) return ++ans,void();
dfs2(d+,num);
REP(i,,S[d].y) num*=S[d].x,dfs2(d+,num);
}
void work() {
init();
int sz = S.size(), now = ;
REP(i,,sz-) {
while (now<B.size()&&B[now].x<S[i].x) C.pb(pii(B[now++].x,));
if (now<B.size()&&B[now].x==S[i].x) {
if (B[now].y>S[i].y) C.pb(S[i]);
++now;
}
}
while (now<B.size()) C.pb(pii(B[now++].x,));
ans = ;
dfs(,m,),dfs2(,);
printf("%lld\n", ans);
} int main() {
gpf[] = ;
REP(i,,N-) if (!gpf[i]) {
for(int j=i;j<N;j+=i) gpf[j]=i;
}
int t;
scanf("%d", &t);
while (t--) work();
}