CF-1154

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E. Two Teams

题意

一个 $1\sim n$ 的排列, 从大到小删数, 当删一个数时将其左右的 $k$ 个数也删了, 若某个数是第奇数次删的输出 1, 否则输出 2

题解

维护两个数组 $l[i],r[i] $ 分别表示第 $i$ 个数左边, 右边的数所在的地方

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 2.1e5;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }

int l[N],r[N],Map[N],ans[N],a[N];

int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n,k;
    in(n,k);
    for1(i,n){
        in(a[i]);
        Map[a[i]]=i;
        l[i]=i-1;
        r[i]=i+1;
    }
    int flag=-1;
    for(int i=n;i>=1;i--){
        if(Map[i]){
            int poi=Map[i],ml=poi,mr=poi;
            for(int i=poi,j=0;i<=n&&j<=k;j++,mr=i=r[i]){
                ans[i]=flag;
                Map[a[i]]=0;
            }
            ml=poi=l[poi];
            for(int i=poi,j=0;i<=n&&j<k;j++,ml=i=l[i]){
                ans[i]=flag;
                Map[a[i]]=0;
            }
            r[ml]=mr;
            l[mr]=ml;
            if(flag==-1)flag=-2;
            else flag=-1;
        }
    }
    for1(i,n)out(-ans[i]);
    puts("");
    return 0;
}

F. Shovels Shop

题意

$n$ 个物品, 其价格分别为 $a_i$ , 有 $m$ 种折扣, 买 $x_i$ 件物品, 便宜的 $y_i$ 件免费, 买 $k(\le 2000)$ 件的最小花费

题解

$dp[i]$ 表示买 $i$ 件物品的最小花费

$dp[i]=min(dp[i-1]+a[i],dp[i-j]+pre[i]-pre[ i-cost[j] ])$

$cost[i]$ 表示买 $i$ 件物品最小需要支付的物品数

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 2.1e5;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }

int a[N],dp[2003],cost[2003],pre[2003];
int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n,m,k,x,y;
    meminf(cost);
    in(n,m,k);
    for1(i,n)in(a[i]);
    for0(i,m){
        in(x,y);
        if(x<2001)
        cost[x]=min(cost[x],x-y);
    }
    sort(a+1,a+n+1);
    for1(i,k)pre[i]=pre[i-1]+a[i];
    for1(i,k){
        dp[i]=dp[i-1]+a[i];
        for1(j,i)
        if(cost[j]<inf)
        dp[i]=min(dp[i],dp[i-j]+pre[i]-pre[i-cost[j]]);
    }
    outln(dp[k]);
    return 0;
}

G. Minimum Possible LCM

题意

给 $n(2\le n \le 1e6)$ 个数, 求最大的 $lcm(a_i,a_j) (1\le a_i\le 1e7)$

题解

枚举 $gcd(a_i,a_j)$ , 找到最小的 $a_i, a_j$

时间复杂度 $O(\frac {10^7}1+\frac {10^7}2+\frac {10^7}3+\cdots+\frac {10^7}{10^7})\approx O(10^7log(10^7))$

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 1.1e7;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }

int a[N][3];

int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n,x,maxx=0;
    in(n);
    for1(i,n){
        in(x);
        maxx=max(maxx,x);
        if(a[x][0]==0){
            a[x][1]=i;
            a[x][0]=1;
        }else if(a[x][0]==1){
            a[x][2]=i;
            a[x][0]=2;
        }
    }
    ll minn=1e18;
    pii ans;
    for1(g,maxx){
        ll x1=0;
        for(int j=g;j<=maxx;j+=g){
            if(x1){
                if(a[j][0]){
                    if(x1*j/g<minn){
                        ans=pii(a[x1][1],a[j][1]);
                        minn=x1*j/g;
                    }
                    break;
                }
            }else{
                if(a[j][0]==1){
                    x1=j;
                }else if(a[j][0]==2){
                    x1=j;
                    if(x1*x1/g<minn){
                        ans=pii(a[x1][1],a[j][2]);
                        minn=x1*j/g;
                    }
                    break;
                }
            }
        }
    }
    if(ans.fi>ans.se)swap(ans.fi,ans.se);
    printf("%d %d\n",ans.fi,ans.se);
    return 0;
}