C題：http://codeforces.com/problemset/problem/407/A

#include<stdio.h> #include<math.h> int main() {int a, b, ax, ay, bx, by;while(~scanf("%d%d",&a, &b)){int flag = 0;for(ax = 1; ax < a; ax++) //枚舉a的橫坐標{for(bx = 1; bx < b; bx++) //枚舉b的橫坐標{ay = (int)sqrt(a*a - ax*ax); //a的縱坐標by = (int)sqrt(b*b - bx*bx); //b的縱坐標if(ax*ax + ay*ay == a*a && bx*bx + by*by == b*b) //兩個都是直角三角形{if(ax*bx == ay*by) //兩個三角形相似{ if(ax != bx) //保證三角形的邊與坐標軸不平行{flag = 1;break;}}}}if(flag)break;}if(!flag)printf("NO\n");else{printf("YES\n");printf("0 0\n");printf("%d %d\n",ax, -ay);printf("%d %d\n",bx, by);}}return 0; }

J題：http://codeforces.com/problemset/problem/405/E

暫時沒有徹底弄懂這個題。

#include<stdio.h> #include<string.h> #include<vector> #include<algorithm> #include<iostream> using namespace std; const int N = 1e5 + 10; struct node {int x, y, z;node(int x, int y, int z) : x(x), y(y), z(z) {} }; vector <int> adj[N]; vector <node> ans; int vis[N]; int tot; int dfs(int u, int fa) {vis[u] = ++tot;//printf("$-> u = %d, fa = %d, vis[%d] = %d\n",u, fa, u, vis[u]);int a = 0;int s = adj[u].size();for(int i = 0; i < s; i++){int v = adj[u][i];if(v == fa) continue;if(!vis[v]){if(dfs(v, u)){// printf("return 1;\n");continue;}if(!a) a = v;else{ans.push_back(node(v, u, a));// printf("****%d %d %d\n",v,u, a);a = 0;}}else if(vis[v] < vis[u]) //判斷v是不是在u之前出現{if(!a) a = v;else{ans.push_back(node(v, u, a));// printf("--->%d %d %d\n",v, u, a);a = 0;}}}if(a){ans.push_back(node(a, u, fa));// printf("####%d %d %d\n",a, u, fa);return 1;}return 0; } int main() {//freopen("a.txt","r",stdin);int n, m, u, v, i, j;while(~scanf("%d%d",&n,&m)){memset(vis, 0, sizeof(vis));memset(adj, 0, sizeof(adj));ans.clear();for(i = 0; i < m; i++){scanf("%d%d",&u,&v);adj[u].push_back(v);adj[v].push_back(u);}if(m&1){printf("No solution\n");continue;}tot = 0;dfs(1, -1);int k = ans.size();for(i = 0; i < k; i++){node tmp = ans[i];printf("%d %d %d\n", tmp.x, tmp.y, tmp.z);}printf("\n");}return 0; } //vis[i]表示第i個點是第幾次訪問到的

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