7-1 压岁钱
不用说
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; int main() { int a, b, c, d; cin >> a >> b >> c >> d; int sum = a + b + c + d; cout << sum; return 0; }
7-2 射击成绩
微米转毫米按环判断。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; int main() { double n; cin >> n; if(n <= 11500 / 2) cout<<10; else if(n <= 27500 / 2) cout<<9; else if(n <= 43500 / 2) cout<<8; else if(n <= 59500 / 2) cout<<7; else if(n <= 75500 / 2) cout<<6; else if(n <= 91500 / 2) cout<<5; else if(n <= 107500 / 2) cout<<4; else if(n <= 123500 / 2) cout<<3; else if(n <= 139500 / 2) cout<<2; else if(n <= 155500 / 2) cout<<1; else cout<<0; return 0; }
7-3 cassels方程
不用说
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; int main() { int n; cin >> n; while (n--) { int x, y, z; cin >> x >> y >> z; if (x * x + y * y + z * z != 3 * x * y * z) cout << "no" << endl; else cout << "yes" << endl; } return 0; }
7-4 相生相克
根据题意相生相克的数字和判断，也可以直接从金到土的数字看另一个数字是啥判断相生还是相克，因为相生相克都是一对一的。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; void judeg(int a, int b) { int sum = a + b; if (sum == 3) cout << "1 ke 2n"; else if (sum == 7) { if (a == 2 || a == 5) cout << "2 ke 5n"; else cout << "3 ke 4n"; } else if (sum == 8) cout << "5 ke 3n"; else if (sum == 5) { if (a == 4 || a == 1) cout << "4 ke 1n"; else cout << "3 sheng 2n"; } else if (sum == 6) { if (a == 2 || a == 4) { cout << "2 sheng 4n"; } else cout << "5 sheng 1n"; } else if (sum == 9) cout << "4 sheng 5n"; else if (a == 1 || a == 3) cout << "1 sheng 3n"; } int main() { int n; cin >> n; while (n--) { int x, y; cin >> x >> y; judeg(x, y); } return 0; }
7-5 7-6太菜了没过
7-5 整除阶乘
对于每个数，直接把n * n + 1对n!的各乘因子求余，最后判断n * n + 1是否变成1来输出结果，用f做是否有结果标记，如果没有就输出none。来自：csdn
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; int main() { int n, m, f = 0; cin >> n >> m; for (int i = n; i <= m; i++) { int sum = i * i + 1; for (int j = 2; j <= i; j++) { if (sum >= j && sum % j == 0) sum /= j; else if (sum < j && j % sum == 0) sum = 1; } if (sum == 1) { cout << i << endl; f = 1; } } if (!f)cout << "none"; return 0; }
7-7 打pta
先判断最后一个字符是否是？，不是直接输出enen，是就从下标2开始判断此下标是否是字符a和前面两个字符是否是t和p，是的话就把flag f设为真，f默认为假，再根据f的真假输出结果。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; int main() { int n; cin >> n; for (int j = 0; j < n; j++) { string s; if (j == 0) getchar(); getline(cin, s); int t = s.length(); int f = 0; if (s[t - 1] != '?') { cout << "enenn"; } else { for (int i = 2; i < t; i++) { if (s[i] == 'a' && s[i - 1] == 't' && s[i - 2] == 'p') { f = 1; break; } } if (f) cout << "yes!n"; else cout << "no.n"; } } return 0; }
7-8 完美对称
从头到尾开始判断区间是否对称，不对称头就顺移到下一位直到找到对称区间，当头等于第一位时就直接完整输出，不是时倒序输出头前面区间。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const ll inf = 2147483647; using namespace std; int main() { int n; cin >> n; vector<int> v(n + 1); for (int i = 1; i <= n; i++) { cin >> v[i]; } for (int i = 1; i <= n; i++) { int k = n, j = i, f = 1; while (k >= j) {//这里判断是否对称 if (v[k] != v[j]) { f = 0; break; } k--; j++; } if (f && i != 1) { cout << v[i - 1]; for (int p = i - 2; p >= 1; p--) { cout << ' ' << v[p]; } break; } else if (f) { cout << v[1]; for (int p = 2; p <= n; p++) { cout << ' ' << v[p]; } break; } } return 0; }
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