思路：想到了用欧拉函数但是开始时没想到欧拉函数的原理就是质因数，浪费了时间，其中还有一些差错：例如以为到3e7最多724个，sqrt(30000000<（5478）导致for循环范围开小，以及数组开小，到最后改完交c++11交超时，过90%的数据，以为是cin的输入问题，还是wa,改成c++14交A了
1 #include <iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<stdlib.h> 5 #include<string.h> 6 #include<map> 7 #include<queue> 8 #include<vector> 9 #include<string> 10 #include<time.h> 11 using namespace std; 12 #define N 30000010 13 int vis[N] = { 0 }; 14 int cnt = 0; 15 int n; 16 long long prime[N]; 17 long long ans=0; 18 void prim() 19 { 20 for (int i = 2; i <= n; i++) 21 { 22 if (!vis[i]) 23 { 24 prime[cnt++] = i; 25 ans += i; 26 } 27 for (int j = 0; j < cnt&&i*prime[j] <= n; j++) 28 { 29 vis[i*prime[j]] = 1; 30 ans += prime[j]; 31 if (i%prime[j] == 0) 32 break; 33 } 34 } 35 printf("%lld", ans); 36 } 37 int main() 38 { 39 scanf("%d", &n); 40 prim(); 41 return 0; 42 }
题解：用sort排序，尾减头就ok
1 #include <iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<stdlib.h> 5 #include<string.h> 6 #include<map> 7 #include<queue> 8 #include<vector> 9 #include<string> 10 #include<time.h> 11 using namespace std; 12 long long ans= 0; 13 int a[100010]; 14 int main() 15 { 16 int n,k; 17 cin >> n >> k; 18 for (int i = 0; i < n; i++) 19 cin >> a[i]; 20 sort(a, a + n); 21 int l = 0, r = n - 1; 22 while (k--) 23 { 24 ans += a[r--] - a[l++]; 25 } 26 cout << ans << endl; 27 return 0; 28 }
题意：注意点没有重复的，且只能改变一个点，所以可以任取两个点进行判断与剩下的两个点的距离相等
1 #include <iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<stdlib.h> 5 #include<string.h> 6 #include<map> 7 #include<queue> 8 #include<vector> 9 #include<string> 10 #include<time.h> 11 using namespace std; 12 struct node 13 { 14 int x, y; 15 }aa,bb,cc,dd,so[7]; 16 vector<node>a, b, c, d; 17 int len(node a, node b) 18 { 19 int x = (a.x - b.x)*(a.x - b.x); 20 int y = (a.y - b.y)*(a.y - b.y); 21 return x + y; 22 } 23 int squre(node a, node b, node c, node d) 24 { 25 int flag = 0; 26 if (len(a, b) == len(d, c) && len(a, c) == len(b, d) && len(a, d) == len(b, c)) 27 flag = 1; 28 return flag; 29 } 30 int t = 0; 31 void Dfs(node a, node b, node c, node d) 32 { 33 so[1] = a; so[2] = b; so[3] = c; so[4] = d; 34 for (int i = 1; i <= 4; i++) 35 { 36 so[i].x -= 1; 37 if (squre(so[1], so[2], so[3], so[4])) 38 t = 1; 39 so[i].x += 2; 40 if (squre(so[1], so[2], so[3], so[4])) 41 t = 1; 42 so[i].x -= 1; 43 } 44 for (int i = 1; i <= 4; i++) 45 { 46 so[i].y -= 1; 47 if (squre(so[1], so[2], so[3], so[4])) 48 t = 1; 49 so[i].y += 2; 50 if (squre(so[1], so[2], so[3], so[4])) 51 t = 1; 52 so[i].y -= 1; 53 } 54 } 55 int main() 56 { 57 cin >> aa.x >> aa.y; 58 cin >> bb.x >> bb.y; 59 cin >> cc.x >> cc.y; 60 cin >> dd.x >> dd.y; 61 if (squre(aa,bb,cc,dd)) 62 { 63 cout << "wen" << endl; 64 return 0; 65 } 66 Dfs(aa, bb, cc, dd); 67 if (t == 1) 68 { 69 cout << "hai xing" << endl; 70 return 0; 71 } 72 cout << "wo jue de bu xing" << endl; 73 return 0; 74 }