回归了回归了！考试周终于过去了。博客也该更新了
Problem A Valera and X
题意：让你判断一个字符方阵是不是x。
思路：很好办暴力就行了不多说了。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-03-19 23:22 5 * Filename : Codeforce_237_2_A.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 1010; 34 char Map[LEN][LEN]; 35 int vis[LEN][LEN]; 36 37 int main() 38 { 39 // freopen("in.txt", "r", stdin); 40 41 int n; 42 while(cin >> n){ 43 memset(vis, 0, sizeof vis); 44 for(int i=0; i<n; i++){ 45 for(int j=0; j<n; j++){ 46 cin >> Map[i][j]; 47 } 48 } 49 char tmp = Map[0][0]; 50 int f = 1; 51 for(int i=0; i<n; i++){ 52 vis[i][i] = vis[i][n-i-1] = 1; 53 if(Map[i][i] != tmp) f = 0; 54 if(Map[i][n-i-1] != tmp) f = 0; 55 if(!f) break; 56 } 57 if(Map[0][1] == tmp) f = 0; 58 tmp = Map[0][1]; 59 for(int i=0; i<n; i++){ 60 for(int j=0; j<n; j++){ 61 if(!vis[i][j] && Map[i][j] != tmp) f = 0; 62 } 63 } 64 if(f) cout << "YES" << endl; 65 else cout << "NO" << endl; 66 } 67 return 0; 68 }
Problem B Marathon
题意：一个坐标轴上一个正方形，右下角为原点，一个人沿着正方形逆时针跑。没跑k距离要喝水，一共和n次，让你输出吗，每次喝水人的坐标。
思路：先摸一下圈长，然后再分四条边讨论。注意这里由于是浮点数所以不能直接模，必须转换一下。注意用longlong就行。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-03-19 23:22 5 * Filename : Codeforce_237_2_B.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 double n, k; 34 35 int calc(double num){ 36 num -= ((ll)(num/(4*n))*4*n); 37 return int(num/n); 38 } 39 40 int main() 41 { 42 // freopen("in.txt", "r", stdin); 43 44 double t; 45 while(scanf("%lf%lf", &n, &k)!=EOF){ 46 scanf("%lf", &t); 47 double st = 0; 48 for(int i=1; i<=t; i++){ 49 int tag = calc(i*k); 50 double x = (i*k) - ((ll)((i*k)/n)*n); 51 if(tag == 0) printf("%lf %lf ", x, 0.0); 52 else if(tag == 1) printf("%lf %lf ", n, x); 53 else if(tag == 2) printf("%lf %lf ", n-x, n); 54 else printf("%lf %lf ", 0.0, n-x); 55 } 56 } 57 return 0; 58 }
Problem C Restore Graph
题意：告诉你图上的定点数，每个点上最大边数，已经选定的点到所有点的最短路（选定的点不知道）。让你构造这幅图。不可能输出-1.
思路：我是按照树的形式来构造的首先距离0的点为根（超过一个则不可能）然后1的点为0的点的孩子以此类推。。。由于每个点（除根之外）只能有k-1个孩子。所以每次判断一下即可。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-03-19 23:22 5 * Filename : Codeforce_237_2_C.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 100000+10; 34 int k, n, d[LEN], top; 35 vector<int> Map[LEN]; 36 37 int main() 38 { 39 // freopen("in.txt", "r", stdin); 40 41 while(scanf("%d%d", &n, &k)!=EOF){ 42 queue<pii> q; 43 while(!q.empty())q.pop(); 44 for(int i=0; i<LEN; i++) Map[i].clear(); 45 top = 0; 46 for(int i=0; i<n; i++){ 47 scanf("%d", &d[i]); 48 Map[d[i]].PB(i); 49 top = max(top, d[i]); 50 } 51 int ans = 1; 52 if(Map[0].size() != 1) ans = 0; 53 else{ 54 for(int i=1; i<=top; i++){ 55 if(Map[i].size() == 0) ans = 0; 56 } 57 for(int i=1; i<=top; i++){ 58 int tp = 0, cnt = 0; 59 if(i==1) cnt--; 60 for(int j=0; j<Map[i].size(); j++){ 61 q.push(MP(Map[i-1][tp]+1, Map[i][j]+1)); 62 cnt++; 63 if(cnt > k-1) { 64 ans = 0; 65 break; 66 } 67 else if(cnt == k-1){ 68 cnt = 0; 69 tp ++; 70 if(tp == Map[i-1].size() && j!=Map[i].size()-1){ 71 ans = 0; 72 break; 73 } 74 } 75 } 76 if(!ans) break; 77 } 78 } 79 if(!ans) printf("-1 "); 80 else{ 81 printf("%d ", q.size()); 82 while(!q.empty()){ 83 pii tmp = q.front();q.pop(); 84 printf("%d %d ", tmp.first, tmp.second); 85 } 86 } 87 } 88 return 0; 89 }