牛客小白月赛16
A、小石的签到题
思路：签到题。简单列举一下即可发现当且仅当 $n == 1$ 时，先手小石必输，其他情况下都必赢！
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 using namespace std; 18 typedef long long LL; 19 typedef unsigned long long ULL; 20 const double eps = 1e-6; 21 const double PI = acos(-1.0); 22 const int maxn = 1e5+5; 23 const int inf = 0x3f3f3f3f; 24 25 int n; 26 27 int main(){ 28 while(cin >> n) { 29 if(n == 1)cout <<"Yang" << endl; 30 else cout << "Shi" <<endl; 31 } 32 return 0; 33 }
B、小雨的三角形
思路：暴力就完事了！+1
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 using namespace std; 18 typedef long long LL; 19 typedef unsigned long long ULL; 20 const double eps = 1e-6; 21 const double PI = acos(-1.0); 22 const int maxn = 1e5+5; 23 const int inf = 0x3f3f3f3f; 24 const LL mod = 1e9+7; 25 26 int n, m, x, y; 27 LL mp[1005][1005], ans[1005]; 28 29 int main(){ 30 memset(mp, 0, sizeof(mp)); 31 mp[1][1] = 1; 32 mp[2][1] = mp[2][2] = 2; 33 for(int i = 3; i <= 1000; ++i) mp[i][1] = mp[i][i]= i; 34 for(int i = 3; i <= 1000; ++i) { 35 for(int j = 2; j < i; ++j) { 36 mp[i][j] = (mp[i - 1][j] + mp[i - 1][j - 1] ) % mod; 37 } 38 } 39 memset(ans, 0, sizeof(ans)); 40 for(int i = 1; i <= 1000; ++i) { 41 for(int j = 1; j <= i; ++j) { 42 ans[i] = (ans[i] + mp[i][j]) % mod; 43 } 44 } 45 while(cin >> n >> m) { 46 for(int i = 1; i <= m; ++i) { 47 cin >> x >> y; 48 LL res = 0; 49 for(int i = x; i <= y; ++i) { 50 res = (res + ans[i]) % mod; 51 } 52 cout << res << endl; 53 } 54 } 55 return 0; 56 }
C、小石的海岛之旅
思路：暴力就完事了！+2
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 using namespace std; 18 typedef long long LL; 19 typedef unsigned long long ULL; 20 const double eps = 1e-6; 21 const double PI = acos(-1.0); 22 const int maxn = 1e5+5; 23 const int inf = 0x3f3f3f3f; 24 25 int n, m, a, cnt, h[1005]; 26 27 int main(){ 28 while(cin >> n >> m) { 29 for(int i = 0; i < n; ++i) cin >> h[i]; 30 for(int i = 1; i <= m; ++i) { 31 cin >> a; 32 cnt = 0; 33 bool flag = false; 34 for(int i = 0; i < n; ++i) { 35 if(a < h[i] && !flag) ++cnt, flag = true; 36 else if(a >= h[i] && flag) flag = false; 37 } 38 cout << cnt << endl; 39 } 40 } 41 return 0; 42 }
D、小阳买水果
思路：前缀和的差即所求答案 $ [k, i] = [1, i] - [1, k) , ( k \leq i) $，关键点：维护一个不增序列a和最长连续子序列的和大于0。因为要求最长连续子序列（区间和大于0）的长度，那么对于第 i 个数，从不增序列a中找到一个小于当前前缀和 $ sum_i $ 的最小值，因为是不增序列，所以可以用二分来查找这个最小值的位置。为什么要维护这么一个不增序列呢？如果能从 $ [1, i - 1] $ 找到，因为从小 ---> 变大，那么所求区间 $ [k , i] $ 中所有元素的和一定是正数，也就是在他满意的条件下即可求出最长连续子序列的长度！
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 #include <bitset> 18 using namespace std; 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左 22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向 23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; 24 const double eps = 1e-6; 25 const double PI = acos(-1.0); 26 const int maxn = 2e6+5; 27 const int inf = 0x3f3f3f3f; 28 29 int n, x, sum, res, lt, rt, ans, mid, a[maxn]; 30 31 int main() { 32 while(~scanf("%d", &n)) { 33 memset(a, 0, sizeof(a)); 34 ans = sum = 0; 35 for(int i = 1; i <= n; ++i) { 36 scanf("%d", &x); 37 sum += x; // 前i个数的前缀和 38 a[i] = min(a[i - 1], sum); // 维护不增序列 39 lt = 0, rt = i - 1, res = -1; 40 while(lt <= rt) { // 从不增序列中找出最小值 41 mid = (lt + rt) >> 1; 42 if(a[mid] >= sum) lt = mid + 1; 43 else rt = mid - 1, res = mid; 44 } 45 if(res != -1) ans = max(ans, i - res); 46 } 47 printf("%d\n", ans); 48 } 49 return 0; 50 }
E、小雨的矩阵
思路：深搜入门题！
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 #include <bitset> 18 using namespace std; 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左 22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向 23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; 24 const double eps = 1e-6; 25 const double PI = acos(-1.0); 26 const int maxn = 1e5+5; 27 const int inf = 0x3f3f3f3f; 28 29 int n, a[10][10]; 30 set<int> syt; 31 32 bool check(int x, int y) { 33 if(x < 0 || y < 0 || x >= n || y >= n) return false; 34 return true; 35 } 36 37 void dfs(int x, int y, int sum) { 38 if(x == n - 1 && y == n - 1) { 39 syt.insert(sum); 40 return; 41 } 42 for(int i = 1; i <= 2; ++i) { 43 int dx = x + dir[i][0], dy = y + dir[i][1]; 44 if(check(dx, dy)) dfs(dx, dy, sum + a[dx][dy]); 45 } 46 } 47 48 int main() { 49 while(cin >> n) { 50 syt.clear(); 51 for(int i = 0; i < n; ++i) { 52 for(int j = 0; j < n; ++j) { 53 cin >> a[i][j]; 54 } 55 } 56 dfs(0, 0, a[0][0]); 57 cout << syt.size() << endl; 58 } 59 return 0; 60 }
G、小石的图形
思路：小学数学，当且仅当绳子围成半圆时面积最大！
AC代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 #include <iomanip> 8 #include <complex> 9 #include <string> 10 #include <vector> 11 #include <set> 12 #include <map> 13 #include <list> 14 #include <deque> 15 #include <queue> 16 #include <stack> 17 using namespace std; 18 typedef long long LL; 19 typedef unsigned long long ULL; 20 const double eps = 1e-6; 21 const double PI = acos(-1.0); 22 const int maxn = 1e5+5; 23 const int inf = 0x3f3f3f3f; 24 const LL mod = 1e9+7; 25 26 int n; 27 28 int main(){ 29 while(cin >> n) { 30 printf("%.3f\n", 0.5 * n * n / PI); 31 } 32 return 0; 33 }
