比较简单的一场。
题目链接：https://codeforces.com/contest/1167
A：
手速快三分钟就切了。
1 #include <bits/stdc++.h>
2 #define ll long long
3 #define pb push_back
4 #define mp make_pair
5 #define sot(a,b) sort(a+1,a+1+b)
6 #define rep(i,a,b) for (int i=a;i<=b;i++)
7 #define eps 1e-8
8 #define int_inf (1<<30)-1
9 #define ll_inf (1LL<<62)-1
10 #define lson curPos<<1
11 #define rson curPos<<1|1
12
13 using namespace std;
14
15 int t;
16
17 int main()
18 {
19 cin>>t;
20 while(t--){
21 int n; string s; cin>>n>>s;
22 int flag=0;
23 for (int i=0;i<n-10;i++) if (s[i]==‘8‘) flag=1;
24 if (flag) puts("YES"); else puts("NO");
25 }
26 return 0;
27 }
B：
非常简单的一道交互。读入a[1]*a[2]，a[2]*a[3]，a[3]*a[4]，a[4]*a[5]之后next_permutation枚举就完事了。
1 /* basic header */
2 #include <bits/stdc++.h>
3 /* define */
4 #define ll long long
5 #define dou double
6 #define pb emplace_back
7 #define mp make_pair
8 #define fir first
9 #define sec second
10 #define sot(a,b) sort(a+1,a+1+b)
11 #define rep1(i,a,b) for(int i=a;i<=b;++i)
12 #define rep0(i,a,b) for(int i=a;i<b;++i)
13 #define repa(i,a) for(auto &i:a)
14 #define eps 1e-8
15 #define int_inf 0x3f3f3f3f
16 #define ll_inf 0x7f7f7f7f7f7f7f7f
17 #define lson curPos<<1
18 #define rson curPos<<1|1
19 /* namespace */
20 using namespace std;
21 /* header end */
22
23 const int maxn = 6;
24 int a[maxn] = {4, 8, 15, 16, 23, 42}, b[6];
25
26 int main()
27 {
28 rep0(i, 0, 4)
29 {
30 printf("? %d %d\n", i + 1, i + 2); fflush(stdout);
31 scanf("%d", &b[i]);
32 }
33 do
34 {
35 if (a[0]*a[1] == b[0] && a[1]*a[2] == b[1] && a[2]*a[3] == b[2] && a[3]*a[4] == b[3])
36 return printf("! %d %d %d %d %d %d\n", a[0], a[1], a[2], a[3], a[4], a[5]), 0;
37 } while (next_permutation(a, a + 6));
38 return 0;
39 }
C：
一眼数据结构题，DSU维护即可。
1 /* basic header */
2 #include <bits/stdc++.h>
3 /* define */
4 #define ll long long
5 #define dou double
6 #define pb emplace_back
7 #define mp make_pair
8 #define fir first
9 #define sec second
10 #define sot(a,b) sort(a+1,a+1+b)
11 #define rep1(i,a,b) for(int i=a;i<=b;++i)
12 #define rep0(i,a,b) for(int i=a;i<b;++i)
13 #define repa(i,a) for(auto &i:a)
14 #define eps 1e-8
15 #define int_inf 0x3f3f3f3f
16 #define ll_inf 0x7f7f7f7f7f7f7f7f
17 #define lson curPos<<1
18 #define rson curPos<<1|1
19 /* namespace */
20 using namespace std;
21 /* header end */
22
23 struct DSU: vector<int>
24 {
25 vector<int>size;
26 DSU(int n): vector<int>(n), size(n, 1)
27 {
28 rep0(i, 0, n) at(i) = i;
29 }
30 int find(int u)
31 {
32 return at(u) == u ? u : at(u) = find(at(u));
33 }
34 void merge(int u, int v)
35 {
36 u = find(u), v = find(v);
37 if (u != v)
38 {
39 at(v) = u;
40 size[u] += size[v];
41 }
42 }
43 };
44
45 int n, m;
46
47 int main()
48 {
49 scanf("%d%d", &n, &m);
50 DSU dsu(n + 1);
51 rep1(i, 1, m)
52 {
53 int sum; scanf("%d", &sum);
54 if (sum)
55 {
56 int u; scanf("%d", &u);
57 rep0(j, 1, sum)
58 {
59 int v; scanf("%d", &v); dsu.merge(u, v);
60 }
61 }
62 }
63 rep1(i, 1, n) printf("%d ", dsu.size[dsu.find(i)]);
64 puts("");
65 return 0;
66 }
D：
是个很简单的贪心，但是有非常巧妙的做法。
1 /* basic header */
2 #include <bits/stdc++.h>
3 /* define */
4 #define ll long long
5 #define dou double
6 #define pb emplace_back
7 #define mp make_pair
8 #define fir first
9 #define sec second
10 #define sot(a,b) sort(a+1,a+1+b)
11 #define rep1(i,a,b) for(int i=a;i<=b;++i)
12 #define rep0(i,a,b) for(int i=a;i<b;++i)
13 #define repa(i,a) for(auto &i:a)
14 #define eps 1e-8
15 #define int_inf 0x3f3f3f3f
16 #define ll_inf 0x7f7f7f7f7f7f7f7f
17 #define lson curPos<<1
18 #define rson curPos<<1|1
19 /* namespace */
20 using namespace std;
21 /* header end */
22
23 int n, x = 0, y = 0;
24 string s;
25
26 int main()
27 {
28 cin >> n >> s;
29 for (auto i : s)
30 if (i == ‘(‘) cout << x, x ^= 1;
31 else cout << y, y ^= 1;
32 cout << endl;
33 return 0;
34 }
E：
暂时没想到怎么做，应该是个数据结构题。
F：
可以线段树但没必要，树状数组秒杀。
1 /* basic header */
2 #include <bits/stdc++.h>
3 /* define */
4 #define ll long long
5 #define dou double
6 #define pb emplace_back
7 #define mp make_pair
8 #define fir first
9 #define sec second
10 #define sot(a,b) sort(a+1,a+1+b)
11 #define rep1(i,a,b) for(int i=a;i<=b;++i)
12 #define rep0(i,a,b) for(int i=a;i<b;++i)
13 #define repa(i,a) for(auto &i:a)
14 #define eps 1e-8
15 #define int_inf 0x3f3f3f3f
16 #define ll_inf 0x7f7f7f7f7f7f7f7f
17 #define lson curPos<<1
18 #define rson curPos<<1|1
19 /* namespace */
20 using namespace std;
21 /* header end */
22
23 const int mod = 1e9 + 7;
24 const int maxn = 5e5 + 10;
25 ll a[maxn], b[maxn], c[maxn], ans = 0;
26 int n;
27
28 void addmod(ll &a, ll b)
29 {
30 a += b;
31 if (a >= mod) a -= mod;
32 if (a < 0) a += mod;
33 }
34
35 void add(int u, int val)
36 {
37 for (; u <= n; u += u & -u) addmod(c[u], val);
38 }
39
40 ll sum(int u)
41 {
42 ll ret = 0;
43 for (; u > 0; u -= u & -u) addmod(ret, c[u]);
44 return ret;
45 }
46
47 int id(int x)
48 {
49 return lower_bound(b, b + n, x) - b + 1;
50 }
51
52 ll solve()
53 {
54 rep1(i, 1, n) c[i] = 0;
55 ll ans = 0;
56 rep0(i, 0, n)
57 {
58 ll curr = sum(id(a[i]));
59 addmod(ans, (a[i] * curr % mod) * (n - i) % mod);
60 add(id(a[i]), i + 1);
61 }
62 return ans;
63 }
64
65 int main()
66 {
67 scanf("%d", &n);
68 rep0(i, 0, n) scanf("%lld", &a[i]), b[i] = a[i];
69 sort(b, b + n);
70 rep0(i, 0, n)
71 addmod(ans, a[i] * (i + 1) % mod * (n - i) % mod);
72 addmod(ans, solve());
73 reverse(a, a + n);
74 addmod(ans, solve());
75 printf("%lld\n", ans);
76 return 0;
77 }
G：
才3个人过的题，再见。
原文地址：https://www.cnblogs.com/JHSeng/p/10874826.html
时间： 2024-10-17 07:31:38