Codeforces Round #493 (Div. 2)

Codeforces Round #493 (Div. 2)

https://codeforces.com/contest/998/my

A

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define IT set<ll>::iterator
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 100005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>pll;
typedef pair<ll,int> pli;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

int n;
struct sair{
    int v,pos;
}a[1005];

bool cmp(sair a,sair b){
    return a.v<b.v;
}

int main(){
    std::ios::sync_with_stdio(false);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i].v;
        a[i].pos=i;
    }
    if(n==1){
        cout<<-1<<endl;
        return 0;
    }
    sort(a+1,a+n+1,cmp);
    if(n==2&&a[1].v==a[2].v) cout<<-1<<endl;
    else{
        cout<<n-1<<endl;
        for(int i=2;i<=n;i++) cout<<a[i].pos<<" ";
    }
}

 

B

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define IT set<ll>::iterator
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 100005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>pll;
typedef pair<ll,int> pli;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

int n,m;
int a[105];
vector<int>ve;

int main(){
    std::ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>a[i];
    int ji=0,ou=0;
    for(int i=1;i<=n;i++){
        if(a[i]%2) ji++;
        else ou++;
        if(ji==ou&&ji!=0){
            if(i!=n){
                ve.pb(abs(a[i]-a[i+1]));
                ji=0,ou=0;
            }
        }
    }
    sort(ve.begin(),ve.end());
    int co=0;
    for(auto i:ve){
        if(i<=m){
            co++;
            m-=i;
        }
        else break;
    }
    cout<<co<<endl;
}

 

C

题意：有一个长度为n的01串，有两种操作，

1.将一个子串翻转，花费x

2.将一个子串中的0变成1，1变成0，花费 y

问把01串变成全1串的最小花费

思路：通过多次模拟可以发现一个规律（其实不用模拟，想一想就可以发现= =），每次翻转都会减少一串全0子串，所以贪心判断即可

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define IT set<ll>::iterator
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 100005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>pll;
typedef pair<ll,int> pli;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

int n;
ll x,y;
string str;

int main(){
    std::ios::sync_with_stdio(false);
    cin>>n>>x>>y>>str;
    int co=0;
    int flag=0;
    for(int i=0;i<str.length();i++){
        if(str[i]=='0') flag=1;
        else if(flag){
            co++;
            flag=0;
        }
    }
    if(flag) co++;
    if(!co) cout<<0;
    else{
        cout<<min((co-1)*x+y,co*y);
    }
}

 

D

题意：罗马数字只有四个符号，分别代表1，5，10，50，问长度为n的罗马数字可以表示几种不同的十进制数

思路：通过打表可以发现，当n<=11时没有规律，当n>11时是公差为49的等差数列

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define IT set<ll>::iterator
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 100005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>pll;
typedef pair<ll,int> pli;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=1e9+7;
const double oula=0.57721566490153286060651209;
using namespace std;

ll a[]={0,4,10,20,35,56,83,116,155,198,244,292};


int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    cin>>n;
    if(n<=11) cout<<a[n]<<endl;
    else cout<<a[11]+49*(n-11);
}

 

E

题意：有一个n*n的矩阵和三种颜色，问有多少种染色方案，使得矩阵中至少一行或者一列颜色一样

思路：这题我只想到n^2的暴力，但显然过不了这题。。。无耻的去看了下题解，这篇写的很详细：https://www.luogu.org/blog/wcx/solution-cf997c

#include<bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define IT set<ll>::iterator
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 1000006
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>pll;
typedef pair<ll,int> pli;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=998244353;
const double oula=0.57721566490153286060651209;
using namespace std;

ll fac[maxn],inv[maxn];

ll ksm(ll a,ll b){
    ll ans=1;
    a%=MOD;
    while(b){
        if(b&1){
            ans=ans*a%MOD;
        }
        a=a*a%MOD;
        b>>=1;
    }
    return ans;
}

ll cal(ll x,ll y){
    return fac[x]*inv[y]%MOD*inv[x-y]%MOD;
}

int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    cin>>n;
    ll ans=0;
    fac[0]=inv[0]=1;
    for(int i=1;i<=n;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=ksm(fac[i],MOD-2);
    }
    for(int i=1;i<=n;i++){
        ans=(ans+ksm(3,(n*(n-i)+i))*ksm(-1,i+1)*cal(n,i)%MOD+MOD)%MOD;
    }
    ans=ans*2%MOD;
    ll tmp=0;
    for(int i=0;i<n;i++){
        ll t=-ksm(3,i);
        tmp=(tmp+cal(n, i)*ksm(-1, i+1)*(ksm(1 + t, n)-ksm(t, n)+MOD)%MOD+MOD)%MOD;
    }
    ans=(ans+tmp*3)%MOD;
    cout<<ans<<endl;
}