Codeforces Round #582 (Div. 3)

题目链接：https://codeforces.com/contest/1213/

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A：

数奇偶个数。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn=110;
21 int n,a[maxn];
22 
23 int main() {
24     scanf("%d",&n);
25     int aa=0,bb=0;
26     rep1(i,1,n){
27         int x; scanf("%d",&x);
28         if (x&1) aa++; else bb++;
29     }
30     printf("%d\n",min(aa,bb));
31     return 0;
32 }

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B：

单调栈维护一下即可。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 150010;
21 int n, a[maxn];
22 
23 int main() {
24     int t; scanf("%d", &t);
25     while (t--) {
26         scanf("%d", &n);
27         rep1(i, 1, n) scanf("%d", &a[i]);
28         stack<int>s;
29         while (!s.empty()) s.pop();
30         s.push(a[n]);
31         int ans = 0;
32         for (int i = n - 1; i > 0; i--) {
33             if (a[i] > s.top()) ans++;
34             else if (a[i] < s.top()) s.push(a[i]);
35         }
36         printf("%d\n", ans);
37     }
38     return 0;
39 }

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C：

没什么可说的数学题。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int main() {
21     int q; scanf("%d", &q);
22     while (q--) {
23         ll n, m; scanf("%lld%lld", &n, &m);
24         if (n < m) {
25             puts("0");
26             continue;
27         }
28         if (n == m) {
29             printf("%lld\n", m % 10);
30             continue;
31         }
32         ll len = n / m, tmp = m, sum = 0;
33         set<int>s; s.clear();
34         while (!s.count(tmp % 10)) {
35             s.insert(tmp % 10);
36             tmp += m;
37         }
38         tmp -= m;
39         for (auto i : s) sum += i;
40         int round = s.size();
41         sum = len / round * sum;
42         for (ll i = len / round * tmp + m; i <= n; i += m) {
43             sum += i % 10;
44         }
45         printf("%lld\n", sum);
46     }
47     return 0;
48 }

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D：

计算每个数对“比它小的能除2到达的数”的贡献，然后每个数取除2次数前k小即可。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 2e5 + 10;
21 int n, k, ans = int_inf;
22 vector<int>v[maxn];
23 
24 int main() {
25     scanf("%d%d", &n, &k);
26     rep1(i, 1, n) {
27         int x, cnt = 0; scanf("%d", &x);
28         do {
29             v[x].pb(cnt);
30             x >>= 1;
31             cnt++;
32         } while (x);
33     }
34     for (int i = 0; i < maxn; i++) {
35         if ((int)v[i].size() < k) continue;
36         sort(v[i].begin(), v[i].end());
37         int tot = 0;
38         for (int j = 0; j < k; j++) tot += v[i][j];
39         ans = min(ans, tot);
40     }
41     printf("%d\n", ans);
42     return 0;
43 }

View Code

E：

显然不可能无解，只有两种构造方式，验证即可。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 3e5 + 10;
21 int n, pos[3] = {0, 1, 2};
22 char s[4], t[4], ans[maxn];
23 
24 int check() {
25     for (int i = 1; i < 3 * n; i++) {
26         if (ans[i - 1] == s[0] && ans[i] == s[1]) return 0;
27         if (ans[i - 1] == t[0] && ans[i] == t[1]) return 0;
28     }
29     return 1;
30 }
31 
32 int main() {
33     scanf("%d", &n);
34     scanf("%s", s);
35     scanf("%s", t);
36     puts("YES");
37     do {
38         int res = 0;
39         for (int k = 0; k < 3; k++)
40             for (int i = 0; i < n; i++)
41                 ans[res++] = 'a' + pos[k];
42         if (check()) {
43             printf("%s\n", ans);
44             return 0;
45         }
46     } while (next_permutation(pos, pos + 3));
47     do {
48         int res = 0;
49         for (int i = 0; i < n; i++)
50             for (int k = 0; k < 3; k++)
51                 ans[res++] = 'a' + pos[k];
52         if (check()) {
53             printf("%s\n", ans);
54             return 0;
55         }
56     } while (next_permutation(pos, pos + 3));
57     return 0;
58 }

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