Codeforces Round #421 (Div. 1) (BCDE)

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+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head

const int N = 1e6+50;
int n, a[N];
int dl[N], dr[N];

int main() {
    scanf("%d",&n);
    REP(i,1,n) scanf("%d",a+i);
    ll ret = 0, ans = 0;
    int L = 0, R = 0;
    REP(i,1,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]!=1) {
                --dl[n-i+a[i]];
                ++dr[n-i+a[i]];
            }
        }
    }
    ans = ret;
    int pos = 0;
    REP(i,1,n-1) {
        ret += R-L;
        R += dr[i];
        L += dl[i];
        if (a[n-i+1]!=1) --R,++L;
        ret -= abs(a[n-i+1]-n-1);
        ret += abs(a[n-i+1]-1);
        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+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head

const int N = 1e6+50;
int gpf[N];
vector<pii> B,C,S;
ll ans,n,m,s;

void solve(vector<pii> &A, int x) {
    while (x!=1) {
        int p = gpf[x], cnt = 0;
        while (x%p==0) 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(2,1));
    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+1,num,z);
    REP(i,1,C[d].y+1) num/=C[d].x;
    dfs(d+1,num,-z);
}
void dfs2(int d, ll num) {
    if (num>n) return;
    if (d==S.size()) return ++ans,void();
    dfs2(d+1,num);
    REP(i,1,S[d].y) num*=S[d].x,dfs2(d+1,num);
}
void work() {
    init();
    int sz = S.size(), now = 0;
    REP(i,0,sz-1) {
        while (now<B.size()&&B[now].x<S[i].x) C.pb(pii(B[now++].x,0));
        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,0));
    ans = 0;
    dfs(0,m,1),dfs2(0,1);
    printf("%lld\n", ans);
}

int main() {
    gpf[1] = 1;
    REP(i,1,N-1) if (!gpf[i]) {
        for(int j=i;j<N;j+=i) gpf[j]=i;
    }
    int t;
    scanf("%d", &t);
    while (t--) work();
}

原文地址：https://www.cnblogs.com/uid001/p/11614218.html