• Codeforces Round #421 (Div. 1) (BC)


    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 '
    '
    #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
    ", ans, pos);
    }
    View Code

    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$xn+ty=2S+tn, 0le tle m$

    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$xt=2S, 1le tle m$

    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 '
    '
    #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
    ", 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();
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/uid001/p/11614218.html
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