2018 Multi-University Training Contest 1 Solution

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
", 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
", ans);
29         }
30         else
31         {
32             puts("-1");
33         }
34     }
35     return 0;
36 }

View Code

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
", ans);
63     }
64     return 0;
65 }

View Code

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
", ans);
66     }
67     return 0;
68 }

View Code

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
", P[i].id, P[i + 1].id, P[i + 2].id);
35         }
36     }
37     return 0;
38 }

View Code

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], " 
"[i == n]);
58     }
59     return 0;
60 }

View Code

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
", 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
", ans);
43     }
44     return 0;
45 }

View Code

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
", 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
", 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 }

View Code

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原文地址：https://www.cnblogs.com/Dup4/p/9555094.html