标签：eof dig reverse eset view hang begin sed 运算
A. Digits Sequence Dividing
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 char s[N]; int n; 6 7 void solve() 8 { 9 if (n == 2 && s[1] >= s[2]) 10 { 11 puts("NO"); 12 return; 13 } 14 else 15 { 16 puts("YES"); 17 puts("2"); 18 printf("%c %s\n", s[1], s + 2); 19 } 20 } 21 22 int main() 23 { 24 int t; scanf("%d", &t); 25 while (t--) 26 { 27 scanf("%d", &n); 28 scanf("%s", s + 1); 29 solve(); 30 } 31 return 0; 32 }
B. Digital root
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 int t; 6 ll k, x; 7 8 int main() 9 { 10 scanf("%d", &t); 11 while (t--) 12 { 13 scanf("%lld%lld", &k, &x); 14 printf("%lld\n", (k - 1) * 9 + x); 15 } 16 return 0; 17 }
C. Brutality
每次连续的一段中如果大于$k个，取最大的k个$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 200010 6 int n, k; 7 ll a[N]; 8 char s[N]; 9 10 int main() 11 { 12 while (scanf("%d%d", &n, &k) != EOF) 13 { 14 for (int i = 1; i <= n; ++i) scanf("%lld", a + i); 15 scanf("%s", s + 1); s[n + 1] = ‘#‘; 16 priority_queue <ll> pq; pq.push(a[1]); 17 ll res = 0; 18 for (int i = 2; i <= n + 1; ++i) 19 { 20 if (s[i] != s[i - 1]) 21 { 22 for (int j = k; !pq.empty() && j; --j) 23 { 24 res += pq.top(); pq.pop(); 25 } 26 while (!pq.empty()) pq.pop(); 27 } 28 pq.push(a[i]); 29 } 30 printf("%lld\n", res); 31 } 32 return 0; 33 }
D. Compression
Solved.
题意：
有$n \cdot n的01矩阵，把矩阵压缩，压缩后的大小为\frac {n}{x}$
思路：
考虑$x | n, 枚举n的因数，在5200范围下，最多有60个$
$然后每次枚举因数n^2判断是否可以 就可以喜提TLE一枚$
$我们考虑用bitset优化$
$我们先处理出b数组第一行，然后接下来x - 1行都可以直接位运算赋值$
$那么这个复杂度就是O(\frac{n^2}{64} + (\frac{n^2}{x}))$
我们发现$< x 的因数不会太多 当因数 > x 的时候 复杂度就接近O(\frac{n^2}{64})$
$然后和a数字异或可以O(\frac{n^2}{64}) 判断合法性$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 5510 5 int n; 6 bitset <N> a[N], b[N], tmp; 7 char s[N][N]; 8 char Hash[210]; 9 int res; 10 11 void change(int x) 12 { 13 a[x].reset(); 14 for (int i = 1, j = 0; i <= n / 4; ++i) 15 { 16 int num = Hash[s[x][i]]; 17 for (int k = 3; k >= 0; --k) 18 a[x][++j] = (num >> k) & 1; 19 } 20 } 21 22 bool solve(int x) 23 { 24 for (int i = 1; i <= n; i += x) 25 { 26 tmp.reset(); 27 for (int j = 1; j <= n; j += x) 28 for (int k = j; k < j + x; ++k) 29 tmp[k] = a[i][j]; 30 for (int j = i; j < i + x; ++j) 31 { 32 b[j] &= 0; 33 b[j] |= tmp; 34 } 35 } 36 for (int i = 1; i <= n; ++i) if ((a[i] ^ b[i]) != 0) 37 { 38 return false; 39 } 40 return true; 41 } 42 43 int main() 44 { 45 for (int i = 0; i <= 9; ++i) Hash[i + ‘0‘] = i; 46 Hash[‘A‘] = 10; 47 Hash[‘B‘] = 11; 48 Hash[‘C‘] = 12; 49 Hash[‘D‘] = 13; 50 Hash[‘E‘] = 14; 51 Hash[‘F‘] = 15; 52 while (scanf("%d", &n) != EOF) 53 { 54 for (int i = 1; i <= n; ++i) 55 scanf("%s", s[i] + 1); 56 for (int i = 1; i <= n; ++i) change(i); 57 res = 1; 58 vector <int> fac; 59 for (int i = 1; i * i <= n; ++i) if (n % i == 0) fac.push_back(i), fac.push_back(n / i); 60 sort(fac.begin(), fac.end()); 61 fac.erase(unique(fac.begin(), fac.end()), fac.end()); 62 reverse(fac.begin(), fac.end()); 63 for (auto it : fac) if (it > 1) 64 { 65 if (solve(it)) 66 { 67 res = it; 68 break; 69 } 70 } 71 printf("%d\n", res); 72 } 73 return 0; 74 }
Educational Codeforces Round 59 Solution
标签：eof dig reverse eset view hang begin sed 运算
原文地址：https://www.cnblogs.com/Dup4/p/10325551.html
