Gym102174 (The 14-th BIT Campus Programming Contest)-蒲公英云

Sloved	A	B	C	D	E	F	G	H	I	J	K	L
7/12	O	.	O	.	O	O	.	O	.	O	O	.

O for passing during the contest
Ø for passing after the contest
! for attempted but failed
· for having not attempted yet

A.两只脑斧(firevolt,0:05,+1)

签到题，读题找规律即可

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long long
int t,n,m,k;
string b;
int main(){ 
    cin>>n;
    for(int i=1;i<=n;i++){ 
        cin>>b;
        if(b[0]=='0'){ 
            cout<<'X';
        }
        else if(b[0]=='1'||b[0]=='3'||b[0]=='5'){ 
            cout<<'E';
        }
        else
            cout<<'I';
    }
}

C 赛尔逵传说(firevolt,0:57,+3)

由于吃果实获得好攻击力和自身攻击力一样，所以吃一个果实相当于多攻击一次，即少被攻击一次，所以将果实全部用在攻击力高的怪物上即为最优解
爆int

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long long
ll t,n,m,k;
struct ac{ 
    ll x,y;
}f[maxn];
bool cmp(ac a,ac b){ 
    return a.y>b.y||(a.y==b.y&&a.x<b.x);
}
int main(){ 
    cin>>n>>k>>m;
    for(ll i=1;i<=n;i++){ 
        cin>>f[i].x>>f[i].y;
    }
    sort(f+1,f+1+n,cmp);
    ll z=0;
    for(ll i=1;i<=n;i++){ 
        ll zz=f[i].x/k;
        if(f[i].x%k==0)
            zz--;
        if(m>0&&zz>=1){ 
            if(m>=zz){ 
                m-=zz;
                continue;
            }
            else{ 
                zz-=m;
                m=0;
            }
        }
        z=z+zz*f[i].y;
    }
    cout<<z<<endl;
}

E 只有一端开口的瓶子(indiewar,01:57,+1)

考虑最多的情况只需要2个栈，这时形成了一个3个桩的汉诺塔，考虑特判只需要1的情况，比较好的做法是直接用一个栈模拟，如果成功那就只需要一个栈（废话）

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) { ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) {  return b?gcd(b,a%b):a;}
const int maxn = 1e5+1000;
int T,n;
int a[maxn];
int b[maxn];
int main(int argc, char const *argv[])
{ 
    cin >> T;
    while(T--)
    { 
        cin >> n;
        rep(i,0,n)
        { 
            cin >> a[i];
        }
        stack<int> st;
        int t = 1;
        rep(i,0,n)
        { 
            st.push(a[i]);
            while(!st.empty() && st.top() == t)
            { 
                t++;
                st.pop();
            }
        }
        if(st.empty())
        { 
            cout << 1 << endl;
        }
        else
        { 
            cout << 2 << endl;
        }
    }
    return 0;
}

F风王之瞳(huhaowen,0:57,+1)

O(n)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
ll n,m,t;
int main()
{ 
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%lld",&t);
    while(t--){ 
        scanf("%lld%lld",&n,&m);
        if(n > m) swap(n,m);
        ll ans = 0;
        for(int i = 1;i <= n;++i){ 
            ans += (n + 1 - i) * (m + 1 - i) * i;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

H目标是成为数论大师(huhaowen,01:59,+9)

暴力枚举-1e5+7到1e5+7（其实5000应该就够了）判断定义域内满足的解
之前只考虑了定义域，这个题目显然还有值域的约束

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
ll n,m,t,a,b;
ll s[N];
int main()
{ 
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%lld",&t);
    while(t--){ 
        ll cnt = 0;
        scanf("%lld%lld",&a,&b);
        for(int i = -N;i < N;++i){ 
            if(i < b) ;
            else if(a * i == (i - b) * (i - b)) s[cnt++] = i;
        }
        printf("%lld\n",cnt);
        for(int i = 0;i < cnt;++i) printf("%lld%c",s[i],i == cnt - 1 ? '\n' : ' ');
    }
    return 0;
}

J金色传说(indiewar,02:30,+2)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=998244353;
ll powmod(ll a,ll b) { ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) {  return b?gcd(b,a%b):a;}
const int maxn = 5e5 + 1000;
ll a[maxn],b[maxn];
int n;
int main(int argc, char const *argv[])
{ 
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    rep(i,1,maxn-100)
    { 
        ll tem = powmod(10,i);
        a[i] = (tem-1) * tem / 2 % mod;
    }
    b[1] = 10;
    b[2] = 100;
    rep(i,3,maxn-100)
    { 
        b[i] = b[i-1] * 10  % mod + b[i-2] * 20 % mod;
        b[i] %= mod;
    }
    int t;
    cin >> t;
    while(t--)
    { 
        cin >> n;
        ll ans = a[n];
        rep(i,1,n-1)//1 -> n-2
        { 
            ans += (a[i] * 2 * b[n-i-1] % mod)%mod;
            ans %= mod;
        }
        cout << ans << endl;
    }
    return 0;
}

K多项式求导(indiewar,0:21,+1)

模拟求导过程即可，比较好写的方法是从前往后更新

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=998244353 ;
ll powmod(ll a,ll b) { ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) {  return b?gcd(b,a%b):a;}
const int maxn = 110;
ll n,k;
ll a[maxn];
int main(int argc, char const *argv[])
{ 
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> k;
    per(i,0,n+1)
    { 
        cin >> a[i];
        //a[i] = a[i] % mod * i % mod;
    }
    rep(i,0,k)
    { 
        rep(j,0,n+1)
        a[j] = a[j+1] % mod * (j+1) % mod;
    }
    cout << 0;
    per(i,0,n)
    cout << " " << a[i];
    cout << endl;
    return 0;
}