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 }