A - Maximum Multiple
题意：给出一个n 找x, y, z 使得$n = x + y +z$ 并且 $n \equiv 0 \pmod x, n \equiv 0 \pmod y, n \equiv 0 \pmod z$ 并且使得 $x \cdot y \cdot z$ 最大
思路：设$a = \frac{n}{x}, b = \frac{n}{y}, c = \frac{n}{z}$ 那么 $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$ 那么我们考虑去凑 a, b, c
两种方案 ${3, 3, 3}$ 或者 ${2, 4, 4}$ 取max
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 typedef long long ll;
6
7 ll n;
8
9 int main()
10 {
11 int t;
12 scanf("%d", &t);
13 while(t--)
14 {
15 scanf("%lld", &n);
16 if(n % 3 == 0)
17 {
18 ll ans = n / 3;
19 ans *= n / 3;
20 ans *= n / 3;
21 printf("%lld\n", ans);
22 }
23 else if(n % 4 == 0)
24 {
25 ll ans = n / 2;
26 ans *= n / 4;
27 ans *= n / 4;
28 printf("%lld\n", ans);
29 }
30 else
31 {
32 puts("-1");
33 }
34 }
35 return 0;
36 }
B - Balanced Sequence
题意：求如何拼接 使得balanced Sequence 最长
思路：首先预处理，使得剩下的串都是 ((( 或者 ))) 或者 )))(((
((( 这种都放左边 ))) 都放右边 目前的问题就是 )))((( 这种在中间按什么顺序放使得答案最大
我们定义 $L_1, R_1$ $L_2, R_2$ 分别为两个串的左括号数量和右括号数量
假如 1 放在 2前面 那么对答案的贡献是 $min(R_1, L_2)$
假如 1 放在 2 后面 那么对答案的贡献是 $min(R_2, L_1)$
比较对答案的贡献，哪个答案贡献大，选哪种放置方式
如果读答案的贡献一样大，那么我们让左括号多的放前面 因为这样对后面的答案贡献大
Dup4:
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6
7 struct node
8 {
9 int l, r;
10 inline node() {}
11 inline node(int l, int r) : l(l), r(r) {}
12 inline bool operator < (const node &b) const
13 {
14 int t1 = min(l, b.r), t2 = min(r, b.l);
15 return t1 > t2 || (t1 == t2 && (l > b.l));
16 }
17 }arr[N];
18
19 int t, n;
20 char s[N];
21
22 int main()
23 {
24 scanf("%d", &t);
25 while (t--)
26 {
27 int L = 0, R = 0, ans = 0, cnt = 0;
28 scanf("%d", &n);
29 for (int i = 1; i <= n; ++i)
30 {
31 scanf("%s", s);
32 int LL = 0, RR = 0;
33 for (int j = 0, len = strlen(s); j < len; ++j)
34 {
35 if (s[j] == ‘(‘)
36 ++LL;
37 else
38 {
39 if (LL)
40 {
41 --LL;
42 ans += 2;
43 }
44 else
45 ++RR;
46 }
47 }
48 if (LL && RR)
49 arr[++cnt] = node(LL, RR);
50 else if (LL) L += LL;
51 else R += RR;
52 }
53 sort(arr + 1, arr + 1 + cnt);
54 for (int i = 1; i <= cnt; ++i)
55 {
56 int LL = arr[i].l, RR = arr[i].r;
57 ans += min(L, RR) * 2;
58 L -= min(L, RR);
59 L += LL;
60 }
61 ans += min(L, R) * 2;
62 printf("%d\n", ans);
63 }
64 return 0;
65 }
XHT:
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6
7 struct node
8 {
9 int l, r;
10 inline node() {}
11 inline node(int l, int r) : l(l), r(r) {}
12 inline bool operator < (const node &b) const
13 {
14 if(l >= r && b.l < b.r)
15 return true;
16 if(l < r && b.l >= b.r)
17 return false;
18 if(l >= r && b.l >= b.r)
19 return r < b.r;
20 if(l < r && b.l < b.r)
21 return l > b.l;
22 }
23 }arr[N];
24
25 int t, n;
26 char s[N];
27
28 int main()
29 {
30 scanf("%d", &t);
31 while (t--)
32 {
33 int ans = 0;
34 scanf("%d", &n);
35 for (int i = 1; i <= n; ++i)
36 {
37 scanf("%s", s);
38 int LL = 0, RR = 0;
39 for (int j = 0, len = strlen(s); j < len; ++j)
40 {
41 if (s[j] == ‘(‘)
42 ++LL;
43 else
44 {
45 if (LL)
46 {
47 --LL;
48 ans += 2;
49 }
50 else
51 ++RR;
52 }
53 }
54 arr[i] = node(LL, RR);
55 }
56 sort(arr + 1, arr + 1 + n);
57 int L = 0;
58 for (int i = 1; i <= n; ++i)
59 {
60 int LL = arr[i].l, RR = arr[i].r;
61 ans += min(L, RR) * 2;
62 L -= min(L, RR);
63 L += LL;
64 }
65 printf("%d\n", ans);
66 }
67 return 0;
68 }
C - Triangle Partition
水.(排序)
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 1e3 + 10;
6
7 struct node{
8 int x, y, id;
9 inline node(){}
10 inline node(int x, int y, int id):x(x), y(y), id(id){}
11 inline bool operator < (const node &b) const
12 {
13 return x == b.x ? y < b.y : x < b.x;
14 }
15 }P[maxn << 2];
16
17 int n;
18
19 int main()
20 {
21 int t;
22 scanf("%d", &t);
23 while(t--)
24 {
25 scanf("%d", &n);
26 for(int i = 1; i <= 3 * n; ++i)
27 {
28 scanf("%d %d", &P[i].x, &P[i].y);
29 P[i].id = i;
30 }
31 sort(P + 1, P + 1 + 3 * n);
32 for(int i = 1; i <= 3 * n; i += 3)
33 {
34 printf("%d %d %d\n", P[i].id, P[i + 1].id, P[i + 2].id);
35 }
36 }
37 return 0;
38 }
D - Distinct Values
按题意模拟即可。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6
7 struct node
8 {
9 int l, r;
10 inline void scan()
11 {
12 scanf("%d%d", &l, &r);
13 }
14 inline bool operator < (const node &r) const
15 {
16 return l < r.l || (l == r.l && this->r > r.r);
17 }
18 }Data[N];
19
20 int t, n, m;
21 int ans[N];
22 bool vis[N];
23 int R;
24
25 int main()
26 {
27 scanf("%d", &t);
28 while (t--)
29 {
30 memset(ans, 0, sizeof ans);
31 R = 0;
32 scanf("%d%d", &n, &m);
33 for (int i = 1; i <= m; ++i) Data[i].scan();
34 sort(Data + 1, Data + 1 + m);
35 for (int i = 1; i <= m; ++i)
36 {
37 int l = Data[i].l, r = Data[i].r;
38 if (R >= r) continue;
39 if (R + 1 < l)
40 for (int j = R + 1; j < l; ++j) ans[j] = 1;
41 memset(vis, false, sizeof vis);
42 for (int j = l; j <= R; ++j) vis[ans[j]] = true;
43 int L = 1;
44 for (int j = max(R + 1, l); j <= r; ++j)
45 {
46 while (vis[L]) ++L;
47 ans[j] = L;
48 vis[L] = true;
49 }
50 R = max(R, r);
51 }
52 while (R < n)
53 {
54 R++;
55 ans[R] = 1;
56 }
57 for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]);
58 }
59 return 0;
60 }
E - Maximum Weighted Matching
留坑。
F - Period Sequence
留坑。
G - Chiaki Sequence Revisited
题意：定义$a_n$ 求 $\sum_{i = 1} ^ {i = n} a_i$
思路：先不考虑$a_1$
我们对每个最后一个2的幂次数处理出它前面有多少个数， 以及这些数的前缀和是多少
比如说处理出
1 2 4 8 16
1 3 7 15 31
1 5 20 76 288
然后给出n计算的时候 按二进制拆分 比如说
27 = 15 + 7 + 3 + 3
定义$F[i] 为 前面i个数的前缀和$
$ans[27] = F[15] + F[7] + 7 * 8 + F[3] + 3 * (8 + 2) + F[3] + 3 * (8 + 2 + 2)$
对于后面，相当于所有数向右偏移
注意取模
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 64
6 #define ll long long
7
8 const ll MOD = (ll)1e9 + 7;
9
10 int t;
11 ll n;
12 ll a[N], b[N], c[N];
13
14 inline void Init()
15 {
16 a[0] = 1;
17 for (int i = 1; i <= 59; ++i) a[i] = a[i - 1] << 1;
18 b[0] = 1;
19 for (int i = 1; i <= 59; ++i) b[i] = (b[i - 1] << 1) + 1;
20 c[0] = 1;
21 for (int i = 1; i <= 59; ++i) c[i] = ((c[i - 1] << 1) % MOD + (a[i - 1] % MOD) * (b[i - 1] % MOD) % MOD + a[i]) %MOD;
22 // for (int i = 0; i <= 59; ++i) printf("%lld %lld %lld\n", a[i], b[i], c[i]);
23 }
24
25 int main()
26 {
27 Init(); scanf("%d", &t);
28 while (t--)
29 {
30 scanf("%lld", &n); --n;
31 ll ans = 1, tmp = 0;
32 for (int i = 59; i >= 0; --i)
33 {
34 while (n >= b[i])
35 {
36 ans = (ans + c[i]) % MOD;
37 ans = (ans + (b[i] % MOD * tmp) % MOD) % MOD;
38 tmp = (tmp + a[i]) % MOD;
39 n -= b[i];
40 }
41 }
42 printf("%lld\n", ans);
43 }
44 return 0;
45 }
H - RMQ Similar Sequence
留坑。
I - Lyndon Substring
留坑。
J - Turn Off The Light
留坑。
K - Time Zone
按题意模拟即可
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 int t, a, b;
6 char s[10];
7
8 inline void work1()
9 {
10 int ans = 0; int flag = 1;
11 if (s[0] == ‘-‘)
12 flag = -1;
13 int len = strlen(s);
14 for (int i = 1; i < len; ++i)
15 ans = ans * 10 + s[i] - ‘0‘;
16 ans *= flag;
17 int gap = ans - 8;
18 a = (a + gap + 24) % 24;
19 printf("%02d:%02d\n", a, b);
20 return;
21 }
22
23 inline void work2()
24 {
25 int A = 0, B = 0; int flag = 1;
26 if (s[0] == ‘-‘)
27 flag = -1;
28 int len = strlen(s), i;
29 for (i = 1; s[i] != ‘.‘; ++i)
30 A = A * 10 + s[i] - ‘0‘;
31 for (++i; i < len; ++i)
32 B = B * 10 + s[i] - ‘0‘;
33 int tot = a * 60 + b;
34 A *= flag, B = B * 6 * flag;
35 int tmptot = A * 60 + B - 480;
36 tot = (tot + tmptot + 24 * 60) % (24 * 60);
37 printf("%02d:%02d\n", tot / 60, tot % 60);
38 return;
39 }
40
41 int main()
42 {
43 scanf("%d", &t);
44 while (t--)
45 {
46 scanf("%d %d UTC%s", &a, &b, s);
47 bool flag = true;
48 for (int i = 0, len = strlen(s); i < len; ++i)
49 if (s[i] == ‘.‘)
50 {
51 flag = false;
52 break;
53 }
54 if (flag) work1();
55 else work2();
56 }
57 return 0;
58 }
原文地址：https://www.cnblogs.com/Dup4/p/9555094.html
时间： 2024-10-09 15:11:54