Codeforces Round 1049 (Div. 2)（A~D）

比赛链接：https://codeforces.com/contest/2140

A.Shift Sort

思路：要交换，我们可以想到第一个不合法的和最后一个不合法的位置进行交换，此时中间位置的值不会被改变。答案为排序后和原序列不同之处/2

#pragma GCC optimize(3, "Ofast", "inline", "unroll-loops")   
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll                            long long 
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"//                           ???????
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
// const ll N = 5e5 + 10;
// const ull p=1145161651561;
ll ksm(ll x, ll y)
{
    ll ans = 1;
    while (y)
    {
        if (y & 1)
        {
            ans = ans * x % mod;
        }
        x = x * x % mod;
        y >>= 1;
    }
    return ans % mod;
}
void fio()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
const ll N = 3e5 + 5;
struct s {
    ll l,r,id;
    friend bool operator<(const s& a, const s& b) {
        if(a.l!=b.l)return a.l<b.l;
        else return a.r<b.r;
    }
};
ll e[4] = { 0,1,0,-1 };
ll d[4] = { 1,0,-1,0 }; 
void solve() { 
    ll n;
    cin>>n;
    string f;
    cin>>f;
    ll ans=0;
    ll cnt=0;
    for(ll i=0;i<f.size();i++){
        if(f[i]=='0')cnt++;
    }
    ll u=0;
    for(ll i=0;i<f.size();i++){
        if(cnt>0&&f[i]=='1')ans++;
        else if(cnt==0&&f[i]=='0')u++;
        cnt--;
    }
    ans=max(abs(u),abs(ans));
    cout<<ans<<endl;
}
int main() {
    fio();
    ll t = 1;
    cin>>t;
    while (t--)solve();
    return 0;
}

B. Another Divisibility Problem

思路：设len(y)为y的长度对应数值，例如28对应100.x#y%(x+y)=0是等价于\((x*len(y)+y)\%(x+y)=0\),显然y%（x+y）=y,x%(x+y)=x,又因为模意义下的乘积等于正常数乘积取模，故只要len(y)%(x+y)=1即可。那么枚举len(y),然后保证长度对应即可，由于x小于y一个数量级，所以必有答案（例如下面代码cc<=n全跳都可以，当然取消这个判断也对！）

#pragma GCC optimize(3, "Ofast", "inline", "unroll-loops")   
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll                            long long 
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"//                           ???????
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
// const ll N = 5e5 + 10;
// const ull p=1145161651561;
ll ksm(ll x, ll y)
{
    ll ans = 1;
    while (y)
    {
        if (y & 1)
        {
            ans = ans * x % mod;
        }
        x = x * x % mod;
        y >>= 1;
    }
    return ans % mod;
}
void fio()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
const ll N = 3e5 + 5;
struct s {
    ll l,r,id;
    friend bool operator<(const s& a, const s& b) {
        if(a.l!=b.l)return a.l<b.l;
        else return a.r<b.r;
    }
};
ll e[4] = { 0,1,0,-1 };
ll d[4] = { 1,0,-1,0 }; 
void solve() { 
    ll n;
    cin>>n;
    ll cc=1;
    for(ll i=1;i<=9;i++){
        cc*=10;
        if(cc<=n)continue;
        else {
            ll dd=cc/10;
            dd=cc-1-n;
            ll kk=dd;
            ll cnt=0;
            while(kk){
                kk/=10;
                cnt++;
            }
            if(dd<1e9&&dd>0&&cnt==i){
                cout<<dd<<endl;
                return ;
            }
        }
    }
}
int main() {
    fio();
    ll t = 1;
    cin>>t;
    while (t--)solve();
    return 0;
}

C.Ultimate Value

思路：一旦后手肯定直接结束游戏了。只考虑变换一次，如果调换对应奇偶位，那么答案要么就是n-1或者n-2,；如果奇偶位互换，只需考虑枚举1-n,如果第i位为偶数位，那么换的是奇数位（设其为x），那么答案就是\(2a[i]+r-2a[x]-x\),那么我们只需维护奇数位\(-2*a[x]-x\)最大即可，同理奇数位换偶数位也是一样的道理（式子有所区别，但本质一样）。

#pragma GCC optimize(3, "Ofast", "inline", "unroll-loops")   
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll                            long long 
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"//                           ???????
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
// const ll N = 5e5 + 10;
// const ull p=1145161651561;
ll ksm(ll x, ll y)
{
    ll ans = 1;
    while (y)
    {
        if (y & 1)
        {
            ans = ans * x % mod;
        }
        x = x * x % mod;
        y >>= 1;
    }
    return ans % mod;
}
void fio()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
const ll N = 3e5 + 5;
struct s {
    ll l,r,id;
    friend bool operator<(const s& a, const s& b) {
        if(a.l!=b.l)return a.l<b.l;
        else return a.r<b.r;
    }
};
ll e[4] = { 0,1,0,-1 };
ll d[4] = { 1,0,-1,0 }; 
void solve() { 
    ll n;
    cin>>n;
    vector<ll>a(n+2);
    for(ll i=1;i<=n;i++)cin>>a[i];
    ll ans=0;
    if(n&1)ans+=n-1;
    else ans+=n-2;
    ll cc=-3e9;
    ll yy=cc;
    ll cnt=0;
    for(ll i=1;i<=n;i++){
        if(i&1)cnt+=a[i];
        else cnt-=a[i];
    }
    ans+=cnt;
    for(ll i=1;i<=n;i++){
        if(i&1){
            ans=max(ans,cnt+yy-2*a[i]+i);
        }
        else ans=max(ans,cnt+cc+2*a[i]+i);
        if(i&1)cc=max(cc,-2*a[i]-i);
        else yy=max(yy,2*a[i]-i);
    }
    cout<<ans<<endl;
}
int main() {
    fio();
    ll t = 1;
    cin>>t;
    while (t--)solve();
    return 0;
}

D.A Cruel Segment's Thesis

思路：

\[基本思想：答案一定可以认为我先去n个线段的右端点，然后选则其中一半让其变为l后，所有r-所有l的值+每个线段r-l \]

\[如果n为偶数，只需计算每个线段变换代价，-l-r，存起来排序后，选择其中最大的n/2个相加 \]

\[如果n为奇数，那么就枚举剔除一个线段后进行两两组合的答案，我们还是-l-r，存起来后排序，从大到小排序，可以发现答案要么就是前n/2个或者n/2+1个-剔除线段的（-l-r）,所有答案取max \]

\[最后只需把所有线段的r-l加上即可 \]

#pragma GCC optimize(3, "Ofast", "inline", "unroll-loops")   
#include<iostream>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#include<vector>
#include<algorithm>
#include<deque>
#include<cctype>
#include<string.h>
#include<math.h>
#include<time.h>
#include<random>
#include<stack>
#include<string>
#include<functional>
#define ll                            long long 
#define ull unsigned long long
#define lowbit(x) (x & -x)
#define endl "\n"//                           ???????
using namespace std;
mt19937 rnd(time(0));
const ll mod = 998244353;
// const ll N = 5e5 + 10;
// const ull p=1145161651561;
ll ksm(ll x, ll y)
{
    ll ans = 1;
    while (y)
    {
        if (y & 1)
        {
            ans = ans * x % mod;
        }
        x = x * x % mod;
        y >>= 1;
    }
    return ans % mod;
}
void fio()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
const ll N = 3e5 + 5;
struct s {
    ll l,r,id;
    friend bool operator<(const s& a, const s& b) {
        if(a.l!=b.l)return a.l<b.l;
        else return a.r<b.r;
    }
};
ll e[4] = { 0,1,0,-1 };
ll d[4] = { 1,0,-1,0 }; 
void solve() { 
    ll n;
    cin>>n;
    vector<pair<ll,ll>>a(n+5);
    ll sum=0;
    ll cnt=0;
    vector<ll>ls(n+2,0);
    for(ll i=1;i<=n;i++){
        auto &[l,r]=a[i];
        cin>>l>>r;
        sum+=r;
        cnt+=r-l;
        ls[i]=-l-r;
    }
    sort(ls.begin()+1,ls.begin()+1+n);
    reverse(ls.begin()+1,ls.begin()+1+n);
    if(!(n&1)){
        ll ans=cnt+sum;
        //cout<<ans<<endl;
        for(ll i=1;i<=n/2;i++){
            ans+=ls[i];//
        }
        cout<<ans<<endl;
    }
    else {
        ll ans=0;
        vector<ll>pre(n+4,0);
        for(ll i=1;i<=n;i++){
            pre[i]=pre[i-1]+ls[i];
        }
        for(ll i=1;i<=n;i++){
            ll uu=sum-a[i].second;//假设不需要
            ll xx=-a[i].first-a[i].second;
            ll zz=n/2;
            if(ls[zz+1]>=xx){
                ans=max(ans,uu+cnt+pre[zz]);
            }
            else {
                ans=max(ans,uu+cnt-xx+pre[zz+1]);
            }
        }
        cout<<ans<<endl;
    }

}
int main() {
    fio();
    ll t = 1;
    cin>>t;
    while (t--)solve();
    return 0;
}

posted @ 2025-09-10 01:19 长皆阅读(252) 评论(0) 收藏举报

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