2020牛客寒假算法基础集训营2

2020牛客寒假算法基础集训营2

C

/*************************************************************************
	> File Name: c.cpp
	> Author: 
	> Mail: 
	> Created Time: 五  2/ 7 17:38:27 2020
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2005;

ll n,p[N],f[N][N];
int main(int argc, char *argv[]) {
    SIS;
    int n;
    cin >> n;
    for(int i = 1;i <= n;i++)cin >> p[i];
    for(int i = f[0][0] = 1;i <= n; i++){
        f[i][0] = f[i - 1][0] * (mod + 1 - p[i]) % mod;
        for(int j = 1; j <= i; j++){
            f[i][j] = (f[i - 1][j] * (mod + 1 - p[i]) + f[i - 1][j - 1] * p[i]) % mod;
        }
    }
    for(int i = 0;i <= n;i++){
        cout << f[n][i] << " ";
    }
    return 0;
}

D

/*************************************************************************
	> File Name: d.cpp
	> Author: 
	> Mail: 
	> Created Time: 五  2/ 7 18:00:34 2020
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long ll;
const int N = 505;
ll ans;
ll x[N],y[N];
bool check(int i, int j, int k) {
  return ((x[j] - x[i]) * (x[k] - x[i])
          + (y[j] - y[i]) * (y[k] - y[i]) < 0)
         && ((x[j] - x[i]) * (y[k] - y[i])
             - (x[k] - x[i]) * (y[j] - y[i]) != 0);
}

int main(int argc, char *argv[]) {
    SIS;
    int n;
    cin >> n;
    for(int i = 1;i <= n; i++)cin >>x[i] >> y[i];
    for(int i = 1;i <= n;i++){
        for(int j = i + 1;j <= n;j ++){
            for(int k = j + 1;k <= n; k++){
                if(check(j,i,k) || check(k,i,j) || check(i,j,k))++ans;
            }
        }
    }
    cout << ans << endl;


    return 0;
}

E

解释一下,平方后得到

\[i + j + 2\sqrt(i*j) == k \]

只有i , j 必然为整数时候才会有解 且是完全平方数

例如n == 4

ans == 4

如下4种情况

(1,1,)

(4,1,)

(1.4,)

(2,2,)

/*************************************************************************
	> File Name: e.cpp
	> Author: 
	> Mail: 
	> Created Time: 五  2/ 7 18:13:30 2020
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long ll;



int main(int argc, char *argv[]) {
    SIS;
    int ans = 0;
    int n;
    cin >> n;
    for(int i = 1;i * i <= n; i++){
        for(int j = 1;j <= i ;j++){
            if(i * i % j == 0){
                ans += 2;
            }
        }
        ans -= 1;
    }
    cout << ans << endl;



    return 0;
}

F

只需要比较a1 + b1 < a2 + b2;

从最大的依次分配即可

/*************************************************************************
	> File Name: f.cpp
	> Author: 
	> Mail: 
	> Created Time: 五  2/ 7 19:27:09 2020
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long ll;
const int N = 2e5 + 5;
vector<int> sa;
vector<int> sb;
struct Node{
    int a,b,id;
    bool operator < (const Node &tmp) const{
        return a + b > tmp.a + tmp.b;
    }
}a[N];

int main(int argc, char *argv[]) {
    SIS;
    int n;
    cin >> n;
    for(int i = 1;i <= n;i++)cin >> a[i].a;
    for(int i = 1;i <= n;i++)cin >> a[i].b,a[i].id = i;
    sort(a + 1,a + 1 + n);
    for(int i = 1;i <= n; i++){
        ((i & 1) ? sa : sb).push_back(a[i].id);
    }

  for (auto x : sa) printf("%d ", x);
  puts("");
  for (auto x : sb) printf("%d ", x);
  return 0;
}