CodeFoces 1373

Educational Codeforces Round 90 (Rated for Div. 2)

A

根据题意模拟一下即可

#include <bits/stdc++.h>
#define sz(X) ((int)(X).size())
#define all(X) (X).begin(), (X).end()
#define rep(I, N) for (int I = 1; I <= (N); ++I)
#define repp(I, N) for (int I = 0; I < (N); ++I)
#define FOR(I, A, B) for (int I = (A); I <= (B); ++I)
#define FORR(I, A, B) for (int I = (A); I >= (B); I--)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define mp make_pair
#define pb push_back
#define ms0(X) memset((X), 0, sizeof((X)))
#define ms1(X) memset((X), -1, sizeof((X)))
#define len(X) strlen(X)
#define f first
#define s second
using namespace std;
const int N = 2e5 + 5;
const double eps = 1e-7;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
ll gcd(ll a,ll b){return b>0?gcd(b,a%b):a;}
ll ksm(ll a,ll b){ll ans = 1;while(b){if(b&1) ans = ans * a % mod;a = a * a % mod;b >>= 1;}return ans % mod;}
int a[N];
std::vector<int> v1,v2;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t--)
    {
       ll a,b,c;
       cin >> a >> b >> c;
       int flag = 0;
       if(a >= c)
       {
          cout << "-1" << ' ' << b <<endl;
       }
       else 
       {
           ll ans1 = a * b;
           if(ans1 <= c)
           cout << "1" << ' ' << "-1" << endl;
           else cout << "1"  << ' ' << b << endl; 
       }
    }
    return 0;
} 

B

比赛时候我 \(5min\) 过的 \(B\) ,就是找到 \(0\) 和 \(1\) 的个数，然后取个数较小的那一个，看一下个数模 \(2\) 的余数。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    string s;
    while(t--)
    {
        cin >> s;
        int ans1= 0 ,ans2= 0;
        for(int i = 0;i < s.size();i++)
        {
            if(s[i] == '0') ans1++;
            else ans2++;
        }
        if(ans1 > ans2) swap(ans1,ans2);
        if(ans1 % 2 == 0)
        {
            cout << "NET" << endl;
        }
        else cout << "DA" << endl;    
    }
    return 0;
} 

C

丑到不行的代码

其实就是 \(+\) 看成 \(+1\) \(-\) 号看成 \(-1\) 。然后取前缀和，找到负数成下降的下标相加即可。

#include <bits/stdc++.h>
#define sz(X) ((int)(X).size())
#define all(X) (X).begin(), (X).end()
#define rep(I, N) for (int I = 1; I <= (N); ++I)
#define repp(I, N) for (int I = 0; I < (N); ++I)
#define FOR(I, A, B) for (int I = (A); I <= (B); ++I)
#define FORR(I, A, B) for (int I = (A); I >= (B); I--)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())

#define pb push_back
#define ms0(X) memset((X), 0, sizeof((X)))
#define ms1(X) memset((X), -1, sizeof((X)))
#define len(X) strlen(X)
#define f first
#define s second
using namespace std;
const int N = 1e6 + 5;
const double eps = 1e-7;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
ll gcd(ll a,ll b){return b>0?gcd(b,a%b):a;}
ll ksm(ll a,ll b){ll ans = 1;while(b){if(b&1) ans = ans * a % mod;a = a * a % mod;b >>= 1;}return ans % mod;}

char a[N];
ll sum[N];
map<ll,ll> mp;
int main() {
   
    int t;
    scanf("%d",&t);
    while(t--)
    {
       mp.clear();
       scanf("%s",a + 1);
       int n = strlen(a + 1);
       for(int i = 1;i <= n;i++)
       {
          if(a[i] == '-')
          sum[i] = sum[i - 1] - 1;
          else sum[i] = sum[i - 1] + 1;
       }
       ll ans = 0;
       int last = -1;
       int cnt = 0;
       for(int i = 1;i <= n;i++)
       {
        if(sum[i] < 0) {
            ans = i;
            break;
        }
       }
       //cout << ans << endl;
       for(int i= 1;i <= n;i++) 
       {
          if(sum[i] >= 0) {
            cnt ++;
            continue;
          }
          else 
          { 
             if(sum[i] <= last)
             {
                last = sum[i];
                mp[sum[i]] = i + 1;
                if(i == n) mp[sum[i]] = i;
             }   
          }
         
       }
       ll cnt1=  0;
       for(auto it = mp.begin();it != mp.end();it++){
         
         ans += it->second;
         cnt1 = max(cnt1,it->second);
       }
       
       ans -= cnt1;
       ans += n;
       if(cnt == n) ans = n;
       cout << ans << endl;
       // int res = 0;
       // for(int i = 0;i <= 100000;i++)
       // {
       //   int cur = i;
       //   int ok = 1;
       //   for(int j = 1;j <= n;j++)
       //   {
       //      res += 1;
       //      if(a[j] == '+') cur += 1;
       //      else cur -= 1;
       //      if(cur < 0)
       //      {
       //          ok = 0;
       //          break;
       //      }
       //   }
       //   //cout << res << endl;
       //   if(ok) break;
       // }
       // cout << res << endl;
    }
    return 0;
} 

D

这题有点东西。

发现翻转其实就是对应的\(1、2，3、4\)相交换。

#include <bits/stdc++.h>
#define sz(X) ((int)(X).size())
#define all(X) (X).begin(), (X).end()
#define rep(I, N) for (int I = 1; I <= (N); ++I)
#define repp(I, N) for (int I = 0; I < (N); ++I)
#define FOR(I, A, B) for (int I = (A); I <= (B); ++I)
#define FORR(I, A, B) for (int I = (A); I >= (B); I--)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())

#define pb push_back
#define ms0(X) memset((X), 0, sizeof((X)))
#define ms1(X) memset((X), -1, sizeof((X)))
#define len(X) strlen(X)
#define f first
#define s second
using namespace std;
const int N = 2e5 + 5;
const double eps = 1e-7;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
ll gcd(ll a,ll b){return b>0?gcd(b,a%b):a;}
ll ksm(ll a,ll b){ll ans = 1;while(b){if(b&1) ans = ans * a % mod;a = a * a % mod;b >>= 1;}return ans % mod;}
//int a[N];
//std::vector<int> v1,v2;
int a[N];
ll sum1[N],sum2[N];
int x[N],y[N];
ll xx[N];
map<ll,ll> mp;
ll MaxSum(int n,ll *a)
{
    ll sum=0,b=0;
    for(int i=1;i<=n;i++)
    {
        if(b<0) b+=a[i];
        else b=a[i];
        if(b<=sum)
            sum=b;
    }
    for(int i =1;i <= n;i++) 
    {
        sum = min(sum,a[i]);
    }
    return sum;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    //int t;
    //scanf("%d",&t);
    int t;
    cin >> t;
    while(t--)
    {
        int tota = 0,totb = 0;
        int n;
        cin >> n;
        for(int i = 1;i <= n;i++)
        {
            sum1[i] = 0;
            sum2[i] = 0;
            xx[i] = 0;
        }
        rep(i,n) cin >> a[i];
        for(int i = 1;i <= n;i+=2)
        {
            x[++tota] = a[i];
        }
        for(int i = 2;i <= n;i+=2)
            y[++totb] = a[i];
        ll ans  = 0;
        int tt = 0;
        for(int i = 1;i <= tota;i++) ans += x[i];
        for(int i = 1;i <= min(tota,totb);i++)
        {
            xx[++tt] = x[i] - y[i];
        } 
        ll cnt1 = MaxSum(tota,xx);
        
        for(int i = 1;i <= tt;i++) xx[i] = 0;
        tt = 0;
        for(int i = 2;i <= tota;i++)
        {
            xx[++tt] = x[i] - y[i - 1];
        }
        ll cnt2 = MaxSum(tota,xx);
        ans += max(abs(cnt1),abs(cnt2));
        cout << ans << endl;
    }
    return 0;
}