A. Yet Another Dividing into Teams
Description
Solution
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <set>
11 #include <stack>
12 #if __cplusplus >= 201103L
13 #include <unordered_map>
14 #include <unordered_set>
15 #endif
16 #include <vector>
17 #define lson rt << 1, l, mid
18 #define rson rt << 1 | 1, mid + 1, r
19 #define LONG_LONG_MAX 9223372036854775807LL
20 #define ll LL
21 using namespace std;
22 typedef long long ll;
23 typedef long double ld;
24 typedef unsigned long long ull;
25 typedef pair<int, int> P;
26 int n, m, k;
27 const int maxn = 1e5 + 10;
28 template <class T>
29 inline T read()
30 {
31 int f = 1;
32 T ret = 0;
33 char ch = getchar();
34 while (!isdigit(ch))
35 {
36 if (ch == '-')
37 f = -1;
38 ch = getchar();
39 }
40 while (isdigit(ch))
41 {
42 ret = (ret << 1) + (ret << 3) + ch - '0';
43 ch = getchar();
44 }
45 ret *= f;
46 return ret;
47 }
48 template <class T>
49 inline void write(T n)
50 {
51 if (n < 0)
52 {
53 putchar('-');
54 n = -n;
55 }
56 if (n >= 10)
57 {
58 write(n / 10);
59 }
60 putchar(n % 10 + '0');
61 }
62 template <class T>
63 inline void writeln(const T &n)
64 {
65 write(n);
66 puts("");
67 }
68 int a[maxn];
69 int main(int argc, char const *argv[])
70 {
71 #ifndef ONLINE_JUDGE
72 freopen("in.txt", "r", stdin);
73 freopen("out.txt", "w", stdout);
74 #endif
75 int t = read<int>();
76 while (t--)
77 {
78 n = read<int>();
79 for (int i = 0; i < n; i++)
80 a[i] = read<int>();
81 sort(a, a + n);
82 int f = 0;
83 for (int i = 1; i < n; i++)
84 if (a[i] == a[i - 1] + 1)
85 ++f;
86 if (f)
87 puts("2");
88 else
89 puts("1");
90 }
91 return 0;
92 }
B. Books Exchange
Description
给出一个置换群，求每个循环的阶
Solution
dfs瞎搜一手
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <set>
11 #include <stack>
12 #if __cplusplus >= 201103L
13 #include <unordered_map>
14 #include <unordered_set>
15 #endif
16 #include <vector>
17 #define lson rt << 1, l, mid
18 #define rson rt << 1 | 1, mid + 1, r
19 #define LONG_LONG_MAX 9223372036854775807LL
20 #define ll LL
21 using namespace std;
22 typedef long long ll;
23 typedef long double ld;
24 typedef unsigned long long ull;
25 typedef pair<int, int> P;
26 int n, m, k;
27 const int maxn = 2e5 + 10;
28 template <class T>
29 inline T read()
30 {
31 int f = 1;
32 T ret = 0;
33 char ch = getchar();
34 while (!isdigit(ch))
35 {
36 if (ch == '-')
37 f = -1;
38 ch = getchar();
39 }
40 while (isdigit(ch))
41 {
42 ret = (ret << 1) + (ret << 3) + ch - '0';
43 ch = getchar();
44 }
45 ret *= f;
46 return ret;
47 }
48 template <class T>
49 inline void write(T n)
50 {
51 if (n < 0)
52 {
53 putchar('-');
54 n = -n;
55 }
56 if (n >= 10)
57 {
58 write(n / 10);
59 }
60 putchar(n % 10 + '0');
61 }
62 template <class T>
63 inline void writeln(const T &n)
64 {
65 write(n);
66 puts("");
67 }
68 int a[maxn], vis[maxn], step;
69 void dfs(int u, int f)
70 {
71 if (a[u] == f)
72 {
73 vis[u] = step;
74 return;
75 }
76 step++;
77 dfs(a[u], f);
78 vis[u] = step;
79 }
80 int main(int argc, char const *argv[])
81 {
82 #ifndef ONLINE_JUDGE
83 freopen("in.txt", "r", stdin);
84 freopen("out.txt", "w", stdout);
85 #endif
86 int t = read<int>();
87 while (t--)
88 {
89 n = read<int>();
90 memset(vis, 0, sizeof(int) * (n + 1));
91 for (int i = 1; i <= n; i++)
92 a[i] = read<int>();
93 for (int i = 1; i <= n; i++)
94 {
95 if (!vis[i])
96 {
97 step = 1;
98 dfs(i, i);
99 }
100 }
101 for (int i = 1; i <= n; i++)
102 printf("%d ", vis[i]);
103 puts("");
104 }
105 return 0;
106 }
C. Good Numbers
Description
给一个数n，找出一个最小的m使得m>=n，且m是3的幂次之和，同一幂次最多出现依次
Solution
十进制分解为三进制，当某一位是2或3(上一位进位而来)时，当前位赋值0，下一位++，注意判断最后进位，以及最后更新位置之前赋0
三进制转回十进制即为答案
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <set>
11 #include <stack>
12 #if __cplusplus >= 201103L
13 #include <unordered_map>
14 #include <unordered_set>
15 #endif
16 #include <vector>
17 #define lson rt << 1, l, mid
18 #define rson rt << 1 | 1, mid + 1, r
19 #define LONG_LONG_MAX 9223372036854775807LL
20 #define ll LL
21 using namespace std;
22 typedef long long ll;
23 typedef long double ld;
24 typedef unsigned long long ull;
25 typedef pair<int, int> P;
26 ull n, m, k;
27 const int maxn = 1e5 + 10;
28 template <class T>
29 inline T read()
30 {
31 int f = 1;
32 T ret = 0;
33 char ch = getchar();
34 while (!isdigit(ch))
35 {
36 if (ch == '-')
37 f = -1;
38 ch = getchar();
39 }
40 while (isdigit(ch))
41 {
42 ret = (ret << 1) + (ret << 3) + ch - '0';
43 ch = getchar();
44 }
45 ret *= f;
46 return ret;
47 }
48 template <class T>
49 inline void write(T n)
50 {
51 if (n < 0)
52 {
53 putchar('-');
54 n = -n;
55 }
56 if (n >= 10)
57 {
58 write(n / 10);
59 }
60 putchar(n % 10 + '0');
61 }
62 template <class T>
63 inline void writeln(const T &n)
64 {
65 write(n);
66 puts("");
67 }
68 ull p[50];
69 void init()
70 {
71 p[0] = 1;
72 for (int i = 1; i < 50; i++)
73 p[i] = p[i - 1] * 3;
74 }
75 vector<int> t3;
76 ull solve()
77 {
78 t3.clear();
79 while (n)
80 {
81 t3.emplace_back(n % 3);
82 n /= 3;
83 }
84 int sz = t3.size();
85 int f = -1;
86 for (int i = 0; i < sz; i++)
87 if (t3[i] == 2 || t3[i] == 3)
88 {
89 f = i;
90 if (i != sz - 1)
91 {
92 t3[i] = 0;
93 t3[i + 1]++;
94 }
95 else
96 {
97 t3[i] = 0;
98 t3.emplace_back(1);
99 }
100 }
101 sz = t3.size();
102 ull res = 0;
103 if (f != -1)
104 for (int i = 0; i < f; i++)
105 t3[i] = 0;
106 for (int i = 0; i < sz; i++)
107 if (t3[i])
108 res += p[i];
109 return res;
110 }
111 int main(int argc, char const *argv[])
112 {
113 #ifndef ONLINE_JUDGE
114 freopen("in.txt", "r", stdin);
115 // freopen("out.txt", "w", stdout);
116 #endif
117 init();
118 int t = read<int>();
119 while (t--)
120 {
121 n = read<ull>();
122 writeln(solve());
123 }
124 return 0;
125 }
D1. Too Many Segments
Description
给n个区间，一个上界k
问最少删除多少区间，使得区间内任一点的覆盖数不大于k
贪心策略，从左往右找每个需要删区间的点，对于区间选择采取满足l<=i且r最大
由于区间和n值较小，O(n^3)亦可过
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#include <vector>
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define LONG_LONG_MAX 9223372036854775807LL
#define ll LL
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> P;
int n, m, k;
const int maxn = 1e5 + 10;
template <class T>
inline T read()
{
int f = 1;
T ret = 0;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (isdigit(ch))
{
ret = (ret << 1) + (ret << 3) + ch - '0';
ch = getchar();
}
ret *= f;
return ret;
}
template <class T>
inline void write(T n)
{
if (n < 0)
{
putchar('-');
n = -n;
}
if (n >= 10)
{
write(n / 10);
}
putchar(n % 10 + '0');
}
template <class T>
inline void writeln(const T &n)
{
write(n);
puts("");
}
struct node
{
int l, r, idx;
node() {}
node(int l, int r, int idx)
{
this->l = l, this->r = r, this->idx = idx;
}
bool operator<(const node &t1) const
{
if (r == t1.r)
return l < t1.l;
return r < t1.l;
}
};
vector<node> vec;
int p[202], del[202];
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
#endif
n = read<int>(), k = read<int>();
int maxx = 0;
for (int i = 1; i <= n; i++)
{
int x = read<int>(), y = read<int>();
maxx = max(maxx, y);
vec.emplace_back(x, y, i);
for (int j = x; j <= y; j++)
++p[j];
}
for (int i = 1; i <= maxx; i++)
{
while (p[i] > k)
{
int rr = 0, delid = -1;
for (int j = 0; j < n; j++)
if (!del[j] && vec[j].l <= i && vec[j].r > rr)
{
rr = vec[j].r;
delid = j;
}
for (int j = vec[delid].l; j <= vec[delid].r; j++)
--p[j];
del[delid] = 1;
}
}
int res = accumulate(del, del + n, 0);
writeln(res);
for (int i = 0; i < n; i++)
if (del[i])
write(i + 1), putchar(' ');
return 0;
}
D2. Too Many Segments
Description
挖坑
E. By Elevator or Stairs?
Description
Solution
简单dp
dp[i][0]表示走路到i层的最小花费，dp[i][1]表示电梯到i层的最小花费。
考虑状态转移
需要注意dp[2]是一开始就确定的，wa了一发
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <set>
11 #include <stack>
12 #if __cplusplus >= 201103L
13 #include <unordered_map>
14 #include <unordered_set>
15 #endif
16 #include <vector>
17 #define lson rt << 1, l, mid
18 #define rson rt << 1 | 1, mid + 1, r
19 #define LONG_LONG_MAX 9223372036854775807LL
20 #define ll LL
21 using namespace std;
22 typedef long long ll;
23 typedef long double ld;
24 typedef unsigned long long ull;
25 typedef pair<int, int> P;
26 int n, m, k;
27 const int maxn = 2e5 + 10;
28 template <class T>
29 inline T read()
30 {
31 int f = 1;
32 T ret = 0;
33 char ch = getchar();
34 while (!isdigit(ch))
35 {
36 if (ch == '-')
37 f = -1;
38 ch = getchar();
39 }
40 while (isdigit(ch))
41 {
42 ret = (ret << 1) + (ret << 3) + ch - '0';
43 ch = getchar();
44 }
45 ret *= f;
46 return ret;
47 }
48 template <class T>
49 inline void write(T n)
50 {
51 if (n < 0)
52 {
53 putchar('-');
54 n = -n;
55 }
56 if (n >= 10)
57 {
58 write(n / 10);
59 }
60 putchar(n % 10 + '0');
61 }
62 template <class T>
63 inline void writeln(const T &n)
64 {
65 write(n);
66 puts("");
67 }
68 ll dp[maxn][2];
69 ll a[maxn], b[maxn], c;
70 int main(int argc, char const *argv[])
71 {
72 #ifndef ONLINE_JUDGE
73 freopen("in.txt", "r", stdin);
74 freopen("out.txt", "w", stdout);
75 #endif
76 n = read<int>(), c = read<int>();
77 for (int i = 1; i < n; i++)
78 a[i] = read<ll>();
79 for (int i = 1; i < n; i++)
80 b[i] = read<ll>();
81 dp[2][1] = c + b[1];
82 dp[2][0] = a[1];
83 for (int i = 3; i <= n; i++)
84 {
85 dp[i][0] = min(dp[i - 1][1], dp[i - 1][0]) + a[i - 1];
86 dp[i][1] = min(dp[i - 1][0] + c, dp[i - 1][1]) + b[i - 1];
87 }
88 for (int i = 1; i <= n; i++)
89 write(min(dp[i][0], dp[i][1])), putchar(' ');
90 return 0;
91 }
F. Maximum Weight Subset
Description
挖坑