Educational Codeforces Round 47

A. Game Shopping

签.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 1010
 5 int n, m, c[N];
 6 queue <int> q;
 7 
 8 int main()
 9 {
10     while (scanf("%d%d", &n, &m) != EOF)
11     {
12         while (!q.empty()) q.pop();
13         for (int i = 1; i <= n; ++i) scanf("%d", c + i);
14         for (int i = 1, x; i <= m; ++i) 
15         {
16             scanf("%d", &x);
17             q.push(x);
18         }
19         int res = 0;
20         for (int i = 1; i <= n && !q.empty(); ++i)
21         {
22             if (q.front() >= c[i])
23             {
24                 q.pop();
25                 ++res;
26             }
27         }
28         printf("%d\n", res);
29     }
30     return 0;
31 }
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B. Minimum Ternary String

签.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 100010
 5 char s[N];
 6 int n, cnt;
 7 
 8 int main()
 9 {
10     while (scanf("%s", s + 1) != EOF)
11     {
12         n = strlen(s + 1);
13         cnt = 0;
14         for (int i = 1; i <= n; ++i) if (s[i] == '1')
15             ++cnt;
16         bool flag = false;
17         for (int i = 1; i <= n; ++i)
18         {
19             if (flag && s[i] != '1')
20                putchar(s[i]);
21             else if (!flag)
22             {
23                 if (s[i] == '0')
24                     putchar(s[i]);
25                 else if (s[i] == '2')
26                 {
27                     flag = true;
28                     while (cnt--) putchar('1');
29                     putchar(s[i]);
30                 }
31             }    
32         }
33         if (cnt > 0) while (cnt--) putchar('1');
34         puts("");
35     }
36     return 0;
37 }
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C. Annoying Present

签.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define db long double
 5 #define ll long long
 6 #define N 100010
 7 int n, m, x[N], d[N];
 8 
 9 ll f(int x)
10 {
11     return 1ll * n * (n + 1) / 2 + (1ll * x * x - 1ll * n * x - x);
12 }
13 
14 int main()
15 {
16     while (scanf("%d%d", &n, &m) != EOF)
17     {
18         ll Max = (ll)-1e18, Min = (ll)1e18;
19         for (int i = 1; i <= n; ++i) 
20         {
21             ll tmp = f(i);
22             Max = max(Max, tmp);
23             Min = min(Min, tmp);
24         }
25         for (int i = 1; i <= m; ++i)
26             scanf("%d%d", x + i, d + i);
27         db res = 0;
28         for (int i = 1; i <= m; ++i)
29         {
30             res += 1ll * n * x[i];
31             if (d[i] < 0)
32                 res += 1ll * d[i] * Min;
33             else 
34                 res += 1ll * d[i] * Max;
35         }
36         printf("%.10Lf\n", res * 1.0 / n);
37     }
38     return 0;
39 }
View Code

 

D. Relatively Prime Graph

签。

$考虑质数的密度很大 并且\phi(p) = p - 1 所以暴力找边即可$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 #define N 100010
 6 #define pii pair <int, int>
 7 int n, m;
 8 vector <pii> res;
 9 int phi[N], prime[N];
10 bool check[N];
11 int tot;
12 
13 void init()
14 {
15     memset(check, false, sizeof check);
16     phi[1] = 1;
17     tot = 0;
18     for (int i = 2; i < N; ++i)
19     {
20         if (!check[i])
21         {
22             prime[++tot] = i;
23             phi[i] = i - 1;
24         }
25         for (int j = 1; j <= tot; ++j)
26         {
27             if (i * prime[j] >= N) break;
28             check[i * prime[j]] = true;
29             if (i % prime[j] == 0)
30             {
31                 phi[i * prime[j]] = phi[i] * prime[j];
32                 break;
33             }
34             else
35                 phi[i * prime[j]] = phi[i] * (prime[j] - 1);
36         }
37     }
38 }
39 
40 int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
41 
42 int main()
43 {
44     init();
45     while (scanf("%d%d", &n, &m) != EOF)
46     {
47         ll sum = 0;
48         for (int i = 2; i <= n; ++i) sum += phi[i];
49         if (m < n - 1 || m > sum) 
50         {
51             puts("Impossible");
52             continue;
53         }    
54         puts("Possible");
55         for (int i = 2; i <= n; ++i) printf("%d %d\n", 1, i);
56         m -= n - 1;
57         vector <int> vec;
58         for (int i = 3; i <= n; ++i) vec.push_back(i);
59         sort(vec.begin(), vec.end(), [](int a, int b) { return phi[a] > phi[b]; });
60         for (auto it : vec)
61         {
62             for (int i = 2; i < it && m > 0; ++i) if (gcd(it, i) == 1)
63             {
64                 printf("%d %d\n", it, i);
65                 --m;
66             }
67             if (m <= 0) break;
68         }
69     }
70     return 0;
71 }
View Code

 

E. Intercity Travelling

签。

统计$a_i$被计算多少次

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 #define N 1000010
 6 const ll p = (ll)998244353;
 7 int n;
 8 ll a[N], Bit[N];
 9 
10 int main()
11 {
12     while (scanf("%d", &n) != EOF)
13     {
14         for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
15         Bit[0] = 1;
16         for (int i = 1; i <= n; ++i) Bit[i] = (Bit[i - 1] * 2) % p;
17         if (n == 1)
18         {
19             printf("%lld\n", a[1]); 
20             continue;
21         }
22         ll res = 0;
23         for (int i = 1; i <= n; ++i)
24         {
25             res = (res + Bit[n - i] * a[i] % p) % p;
26             if (n - i - 1 >= 0)
27                 res = (res + 1ll * (n - i) * Bit[n - i - 1] % p * a[i] % p) % p;
28         }
29         printf("%lld\n", res);
30     }
31     return 0;
32 }
View Code

 

posted @ 2019-02-18 16:24  Dup4  阅读(130)  评论(0)  编辑  收藏  举报