2019CCPC-江西省赛部分题解

D: Wave

对里面的每个数字的位置进行查找

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(3)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
vector vec[105];
int main(){
  fio;
  int n, c, x;
  cin >> n >> c;
  for(int i = 1; i <= n; ++i) {
    cin >> x;
    vec[x].push_back(i);
  } 
  int ans = 0;
  for(int i = 1; i <= c; ++i) {
    for(int j = 1; j <= c; ++j) {
      if(i == j) continue;
      int len1 = vec[i].size();
      int len2 = vec[j].size();
      int st1 = 0, st2 = 0, temp = 0, sum = 0;
      while(1) {
        while(st1 < len1 && vec[i][st1] < temp) {
          st1 += 1;
        }
        if(st1 == len1) break;
        temp = vec[i][st1];
        sum += 1;
        while(st2 < len2 && vec[j][st2] < temp) {
          st2 += 1;
        }
        if(st2 == len2) break;
        sum += 1;
        temp = vec[j][st2];
      }
      ans = max(ans, sum);
    }
  }
  cout << ans << endl;
  return 0; 
}

F: String

乘法原理

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(6)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
int a[26];
int main(){
  fio;
  int n;
  string dest;
  while(cin >> n) {
    memset(a, 0, sizeof(a));
    cin >> dest;
    for(int i = 0; i < n; ++i) {
      a[dest[i] - 'a'] += 1;
    }
    if(!a[0] || !a['v' - 'a'] || !a['i' - 'a'] || !a['n' - 'a']) {
      cout << "0/1" << endl;
    } else {
      int t = pow(n, 4), k = 1, gcd;
      k = a[0] * a['v' - 'a'] * a['i' - 'a'] * a['n' - 'a'];
      gcd = __gcd(k, t);
      cout << k / gcd << "/" << t / gcd << endl;
    }
  }
  return 0; 
}

G: Traffic

东西方向的先走，然后在南北里面找是不是有时间冲突，有的话就等待。

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(3)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
int a[1005];
int b[1005];
int vis[1005];
int main(){
  fio;
  int n, m;
  while(cin >> n >> m) {
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < n; ++i) cin >> a[i], vis[a[i]] = 1;
    for(int i = 0; i < m; ++i) cin >> b[i];
    for(int i = 0; ; ++i) {
      int flag = 0;
      for(int j = 0; j < m; ++j) {
        if(vis[b[j] + i] == 1) {
          flag = 1;
          break;
        }
      }
      if(flag == 0) {
        cout << i << endl;
        break;
      }
    }
  }
  return 0; 
}

H: Rng

区间概率没搞懂咋算的，但是记得好像是个结论题。

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(6)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
LL quick_pow(LL a, LL b) {
  LL sum = 1;
  while(b) {
    if(b & 1) sum = sum * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return sum;
}
int main(){
  fio;
  LL n;
  while(cin >> n) {
    cout << (n + 1) * quick_pow(2 * n, mod - 2) % mod << endl;
  }
  return 0; 
}
/*
结论题
(n + 1) / 2n
用逆元处理一下 
*/ 

I: Budget

第三位小数有两种情况：
t <= 4 sum += -t * 0.001
t >= 5 sum += (10 - t) * 0.001

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(3)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
int main(){
  fio;
  int n;
  while(cin >> n) {
    double sum = 0.0;
    for(int i = 0; i < n; ++i) {
      string dest;
      int k = 0;
      cin >> dest;
      for(int j = 0; j < dest.size(); ++j) {
        if(dest[j] == '.') {
          k = j;
          break;
        }
      }
      int t = dest[k + 3] - '0';
      if(t <= 4) {
        sum += -t * 0.001;
      } else {
        sum += (10 - t) * 0.001;
      }
    }
    cout << fix << sum << endl;
  }
  return 0; 
}

J: Worker

按比例分配就可以了。

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(6)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
LL a[1005];
int main(){
  fio;
  LL n, m;
  while(cin >> n >> m) {
    LL lcm = 1, sum = 0;
    for(int i = 0; i < n; ++i) {
      cin >> a[i];
      lcm = lcm * a[i] / __gcd(lcm, a[i]);
    }
    for(int i = 0; i < n; ++i) {
      sum += lcm / a[i];
    }
    if(m % sum) {
      cout << "No" << endl;
    } else {
      cout << "Yes" << endl;
      for(int i = 0; i < n; ++i) {
        cout << m / sum * lcm / a[i];
        if(i != n - 1) cout << " ";
        else cout << endl;
      }
    }
  }
  return 0; 
}

K: Class

#include
#include
#include
#include
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(6)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
int main(){
  fio;
  int x, y, a, b;
  cin >> x >> y;
  a = (x + y) / 2;
  b = (x - y) / 2;
  cout << a * b << endl;
  return 0; 
}