Educational Codeforces Round 74 #div2 ABCD

A. Prime Subtraction

Description

给两个数$x,y,x \gt y$,判断$x-y$是否为一个素数的整数倍

Solution

由唯一分解可得每个大于1的数都可以唯一分解为素数幂次之积，显然只需要判断x-y是否大于1即可

 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 <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(‘ ‘)
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 1e5 + 10;
31 template <class T>
32 inline T read()
33 {
34     int f = 1;
35     T ret = 0;
36     char ch = getchar();
37     while (!isdigit(ch))
38     {
39         if (ch == ‘-‘)
40             f = -1;
41         ch = getchar();
42     }
43     while (isdigit(ch))
44     {
45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
46         ch = getchar();
47     }
48     ret *= f;
49     return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54     if (n < 0)
55     {
56         putchar(‘-‘);
57         n = -n;
58     }
59     if (n >= 10)
60     {
61         write(n / 10);
62     }
63     putchar(n % 10 + ‘0‘);
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68     write(n);
69     puts("");
70 }
71 int main(int argc, char const *argv[])
72 {
73 #ifndef ONLINE_JUDGE
74     freopen("in.txt", "r", stdin);
75     // freopen("out.txt", "w", stdout);
76 #endif
77     int t = read<int>();
78     while (t--)
79     {
80
81         ll x = read<ll>(), y = read<ll>();
82         if (x - y == 1)
83             puts("NO");
84         else
85             puts("YES");
86     }
87
88     return 0;
89 }

B. Kill ‘Em All

Description

小明想要消灭一群在x正半轴的怪兽。

给出一个怪兽x坐标序列，导弹作用半径r。

小明每次可以选择一个坐标抛一个炸弹，处于炸弹中心的怪兽直接被消灭，处于炸弹右边的怪兽被推移到x+r的位置，同理左边被推到x-r,x为怪兽坐标。

同样被移到原点或者负半轴也会死亡。

求最少能消灭所有怪兽的炸弹数。

Solution

对于炸弹右边的怪兽，由于右边没有限制，可以被推送到无穷远处，而之后还是需要一颗炸弹来消灭，所以往右推移是不可取的。

那么贪心思路就是从最右开始扔炸弹，知道全部被消灭或者移到非正半轴

  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 <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(‘ ‘)
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 1e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == ‘-‘)
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar(‘-‘);
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + ‘0‘);
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 vector<ll> vec;
 72 int main(int argc, char const *argv[])
 73 {
 74 #ifndef ONLINE_JUDGE
 75     freopen("in.txt", "r", stdin);
 76     // freopen("out.txt", "w", stdout);
 77 #endif
 78     int t = read<int>();
 79     while (t--)
 80     {
 81         n = read<int>();
 82         ll r = read<int>();
 83         vec.clear();
 84         for (int i = 0; i < n; i++)
 85         {
 86             ll x = read<ll>();
 87             vec.emplace_back(x);
 88         }
 89         sort(vec.begin(), vec.end());
 90         auto curend = unique(vec.begin(), vec.end());
 91         int sz = (int)(curend - vec.begin());
 92         int q = 0;
 93         for (int i = sz - 1; i >= 0; i--)
 94         {
 95             if (vec[i] - q * r <= 0)
 96                 break;
 97             ++q;
 98         }
 99         writeln(q);
100     }
101     return 0;
102 }

C. Standard Free2play

Description

Solution

找规律可以发现对于连续的一个1状态，如果其长度为奇数，那么这一段1可以无需消耗钻石安稳降落。

而如果长度为偶数，那么它一定得降到最后一个1之上，到最后一个1的时候由于开关触碰，会掉到x-1位置，这时候需要消耗一个钻石保证认为无伤。

模拟即可。注意特判最后一段1序列

  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 <numeric>
 10 #include <queue>
 11 #include <set>
 12 #include <stack>
 13 #if __cplusplus >= 201103L
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #endif
 17 #include <vector>
 18 #define lson rt << 1, l, mid
 19 #define rson rt << 1 | 1, mid + 1, r
 20 #define LONG_LONG_MAX 9223372036854775807LL
 21 #define pblank putchar(‘ ‘)
 22 #define ll LL
 23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
 24 using namespace std;
 25 typedef long long ll;
 26 typedef long double ld;
 27 typedef unsigned long long ull;
 28 typedef pair<int, int> P;
 29 int n, m, k;
 30 const int maxn = 2e5 + 10;
 31 template <class T>
 32 inline T read()
 33 {
 34     int f = 1;
 35     T ret = 0;
 36     char ch = getchar();
 37     while (!isdigit(ch))
 38     {
 39         if (ch == ‘-‘)
 40             f = -1;
 41         ch = getchar();
 42     }
 43     while (isdigit(ch))
 44     {
 45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
 46         ch = getchar();
 47     }
 48     ret *= f;
 49     return ret;
 50 }
 51 template <class T>
 52 inline void write(T n)
 53 {
 54     if (n < 0)
 55     {
 56         putchar(‘-‘);
 57         n = -n;
 58     }
 59     if (n >= 10)
 60     {
 61         write(n / 10);
 62     }
 63     putchar(n % 10 + ‘0‘);
 64 }
 65 template <class T>
 66 inline void writeln(const T &n)
 67 {
 68     write(n);
 69     puts("");
 70 }
 71 unordered_map<int, int> mp;
 72 int a[maxn];
 73 int main(int argc, char const *argv[])
 74 {
 75 #ifndef ONLINE_JUDGE
 76     freopen("in.txt", "r", stdin);
 77     // freopen("out.txt", "w", stdout);
 78 #endif
 79     int t = read<int>();
 80     while (t--)
 81     {
 82         int h = read<int>();
 83         n = read<int>();
 84         for (int i = 0; i < n; i++)
 85             a[i] = read<int>();
 86         int res = 0;
 87         int cnt = 1;
 88         for (int i = 1; i < n; i++)
 89         {
 90             if (a[i] == a[i - 1] - 1)
 91                 ++cnt;
 92             else
 93             {
 94                 if (!(cnt & 1))
 95                     ++res;
 96                 cnt = 0;
 97             }
 98         }
 99         if (!(cnt & 1) && a[n - 1] > 1)
100             ++res;
101         writeln(res);
102     }
103     return 0;
104 }

D. AB-string

Description

Solution

题解也太妙了8，正着想了好久没想到怎么做。

题解思路，最终good=all-bad

对于bad串，只会出现四种情况。

ABBBBB,BBBBBA

AAAAAB,BAAAAA

那么前后各扫一遍即可，前后可能有长度为2的bad串重复计算，需要减去重复贡献。

 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 <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(‘ ‘)
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 3e5 + 10;
31 template <class T>
32 inline T read()
33 {
34     int f = 1;
35     T ret = 0;
36     char ch = getchar();
37     while (!isdigit(ch))
38     {
39         if (ch == ‘-‘)
40             f = -1;
41         ch = getchar();
42     }
43     while (isdigit(ch))
44     {
45         ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
46         ch = getchar();
47     }
48     ret *= f;
49     return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54     if (n < 0)
55     {
56         putchar(‘-‘);
57         n = -n;
58     }
59     if (n >= 10)
60     {
61         write(n / 10);
62     }
63     putchar(n % 10 + ‘0‘);
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68     write(n);
69     puts("");
70 }
71 char s[maxn];
72 int main(int argc, char const *argv[])
73 {
74 #ifndef ONLINE_JUDGE
75     freopen("in.txt", "r", stdin);
76     // freopen("out.txt", "w", stdout);
77 #endif
78     n = read<int>();
79     scanf("%s", s);
80     ll res = 1LL * (n - 1) * n / 2;
81     int pre = 0;
82     for (int i = 1; i < n; i++)
83         if (s[i] != s[i - 1])
84             res -= i - pre - 1, pre = i;
85     pre = n - 1;
86     for (int i = n - 2; i >= 0; i--)
87         if (s[i] != s[i + 1])
88         {
89             res -= pre - i;
90             pre = i;
91         }
92     writeln(res);
93     return 0;
94 }

原文地址：https://www.cnblogs.com/mooleetzi/p/11754165.html

时间： 2024-07-28 23:35:41

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