第一題：十六進制

輸入一個十進制正整數，輸出這個數的十六進制表示，其中十六進制的10-15用大寫字母A-F表示。

不斷取余，將10-15的數替換成相應大寫字母即可。

#include<iostream> #include<string> #include<cmath> #include<bitset> #include<algorithm> using namespace std; void D_TO_X(int x) {int i, j, arr[16];for (i = 0; x != 0; i++){arr[i] = x % 16;if (arr[i] <= 9){arr[i] += '0'; }else{switch (arr[i]){case 10:arr[i] = 'A'; break;case 11:arr[i] = 'B'; break;case 12:arr[i] = 'C'; break;case 13:arr[i] = 'D'; break;case 14:arr[i] = 'E'; break;case 15:arr[i] = 'F'; break;}}x /= 16;}for (j = i - 1; j >= 0; j--){cout.put(arr[j]);}cout << endl; }int main() {int n;cin >> n;while (n--){int k;cin >> k;D_TO_X(k);}return 0; }

第二題：畫矩形

只有一組案例。先是兩個正整數m和n（3<=m,n<=10），然后是一個字符c，最后是一個0或者1的整數d。m代表矩形圖案的高，n代表矩形圖案的寬，c是用來畫矩形的符號，如果d是0則畫空心的矩形，如果d是1則畫實心的矩形。

比較基礎，換成1與0兩種情況，逐行判斷輸出即可。

#include<iostream> #include<string> #include<cmath> #include<bitset> #include<algorithm> using namespace std;int main() {int m, n;char k;bool x;cin >> m >> n >> k >> x;if (x){for (int i = 0; i < m; i++){for (int i = 0; i < n; i++){cout << k;}cout << endl;}}else{for (int i = 0; i < m; i++){if (i == 0 || i == m - 1){for (int i = 0; i < n; i++){cout << k;}cout << endl;}else{cout << k;for (int i = 0; i < n-2; i++){cout << " ";}cout << k << endl;}}}return 0; }

第三題：從前有個函數

從前有個數學函數f(x)，功能是計算x的因數的個數，例如f(8)=4。（因為8的因數有1、2、4、8）后來有人覺得算一次f函數不過癮，又嵌套了一層，于是變成了f(f(x))，那么f(f(8))=f(4)=3。隨后，就有了更多層的嵌套函數f。人們發現這樣寫太費事了，于是把嵌套層數當成了f函數的第二個參數，例如f(f(8))寫成了f(8,2)，f(9)寫成了f(9,1)。現在，需要計算f(a,b)=?

這里題目輸入的是十億以內的數，可見要節約時間。于是判素后直接取2，在sqrt下*2，完全平方數再加一。

#include<iostream> #include<algorithm> #include<cmath> #include<bitset> #include<string> using namespace std; bool f(int m) {for (int i = 2; i <= sqrt(m); i++){if (m%i == 0)return 0;}if (m>1)return 1;else return 0; }int main() {int n, m, p;cin >> n;while (n--){cin >> m >> p;while (p--){if (f(m)){m = 2; break;}if (m == 1)break;int sum = 0;for (int i = 1; i <= sqrt(m); i++){if (i*i == m)sum++;else if (m%i == 0)sum += 2;}m = sum;}cout << m << endl;} }

第四題：五毒教主吃蘋果

有一天，邪惡的巫師送給了五毒教主一筐蘋果，里面有一些正常的蘋果和一些有毒的蘋果。凡人吃毒蘋果輕則吐血，重則暴斃，然而五毒教主吃一個毒蘋果后只會昏睡一天，并無大礙。昏睡的那天五，毒教主不吃蘋果。例如，第1天吃了一個毒蘋果，則第2天昏睡不吃，第3天繼續吃。
恰逢五毒教主最近減肥，每天最多只吃1個蘋果，直到吃完這一筐蘋果為止。
問：有多少種不同的吃法？（如果有一個正常的蘋果和一個毒蘋果，那么先吃正常蘋果再吃毒蘋果，和先吃毒蘋果再吃正常蘋果，屬于2種不同的吃法。）

看起來很嚇人，實際上就是遞歸。普通的蘋果和毒蘋果就是兩種東西，日期無所謂。

#include<iostream> #include<string> #include<cmath> #include<bitset> #include<algorithm> using namespace std; int t(int a, int b) {if (a == 0 && b == 0)return 1;else if (b == 0)return 1;else if (a == 0)return 1;else return t(a - 1, b) + t(a, b - 1); } int main() {int f;cin >> f;while (f--){int m, n;cin >> m >> n;cout << t(m, n) << endl;}return 0; }

第五題：算出18

有3個非負的整數，想看看是否能夠通過加減乘除運算，使得答案為18。
每個數字必須都用上，且僅能各使用一次。不要求按照順序使用，可以添加上適當的括號以改變計算順序。
注意：除法是按照數學上的運算結果。

這道題有兩種做法，一種是暴力，一種還是遞歸。要考慮到0的情況和是否用到所有數，是否有浮點數誤差，還有數的順序可以替換問題。

#include<iostream> #include<string> #include<cmath> #include<bitset> #include<algorithm> using namespace std;int main() {int f;cin >> f;while (f--){double x1, x2, x3;cin >> x1 >> x2 >> x3;double a = x1; double b = x2; double c = x3;int d = 0;if (a + b + c == 18)d = 1;else if (a + b - c == 18)d = 1;else if (a + b * c == 18)d = 1;else if (c!=0&&a + b / c == 18)d = 1;else if ((a + b) * c == 18)d = 1;else if (c != 0 && (a + b) / c == 18)d = 1;else if (a - b + c == 18)d = 1;else if (a - b - c == 18)d = 1;else if (a - b * c == 18)d = 1;else if (c != 0 && a - b / c == 18)d = 1;else if ((a - b) * c == 18)d = 1;else if (c != 0 && (a - b) / c == 18)d = 1;else if (a * b + c == 18)d = 1;else if (a * b - c == 18)d = 1;else if (a * b * c == 18)d = 1;else if (c != 0 && a * b / c == 18)d = 1;else if (a * (b + c) == 18)d = 1;else if (a * (b - c) == 18)d = 1;else if (a / b + c == 18)d = 1;else if (a / b - c == 18)d = 1;else if (a / b * c == 18)d = 1;else if (c != 0 &&b!=0&& a / b / c == 18)d = 1;else if (b+c!=0&&a / (b + c) == 18)d = 1;else if (b-c!=0&&a / (b - c) == 18)d = 1;a = x2; b = x1; c = x3;if (a + b + c == 18)d = 1;else if (a + b - c == 18)d = 1;else if (a + b * c == 18)d = 1;else if (c != 0 && a + b / c == 18)d = 1;else if ((a + b) * c == 18)d = 1;else if (c != 0 && (a + b) / c == 18)d = 1;else if (a - b + c == 18)d = 1;else if (a - b - c == 18)d = 1;else if (a - b * c == 18)d = 1;else if (c != 0 && a - b / c == 18)d = 1;else if ((a - b) * c == 18)d = 1;else if (c != 0 && (a - b) / c == 18)d = 1;else if (a * b + c == 18)d = 1;else if (a * b - c == 18)d = 1;else if (a * b * c == 18)d = 1;else if (c != 0 && a * b / c == 18)d = 1;else if (a * (b + c) == 18)d = 1;else if (a * (b - c) == 18)d = 1;else if (a / b + c == 18)d = 1;else if (a / b - c == 18)d = 1;else if (a / b * c == 18)d = 1;else if (c != 0 && b != 0 && a / b / c == 18)d = 1;else if (b + c != 0 && a / (b + c) == 18)d = 1;else if (b - c != 0 && a / (b - c) == 18)d = 1;a = x1; b = x3; c = x2;if (a + b + c == 18)d = 1;else if (a + b - c == 18)d = 1;else if (a + b * c == 18)d = 1;else if (c != 0 && a + b / c == 18)d = 1;else if ((a + b) * c == 18)d = 1;else if (c != 0 && (a + b) / c == 18)d = 1;else if (a - b + c == 18)d = 1;else if (a - b - c == 18)d = 1;else if (a - b * c == 18)d = 1;else if (c != 0 && a - b / c == 18)d = 1;else if ((a - b) * c == 18)d = 1;else if (c != 0 && (a - b) / c == 18)d = 1;else if (a * b + c == 18)d = 1;else if (a * b - c == 18)d = 1;else if (a * b * c == 18)d = 1;else if (c != 0 && a * b / c == 18)d = 1;else if (a * (b + c) == 18)d = 1;else if (a * (b - c) == 18)d = 1;else if (a / b + c == 18)d = 1;else if (a / b - c == 18)d = 1;else if (a / b * c == 18)d = 1;else if (c != 0 && b != 0 && a / b / c == 18)d = 1;else if (b + c != 0 && a / (b + c) == 18)d = 1;else if (b - c != 0 && a / (b - c) == 18)d = 1;a = x2; b = x3; c = x1;if (a + b + c == 18)d = 1;else if (a + b - c == 18)d = 1;else if (a + b * c == 18)d = 1;else if (c != 0 && a + b / c == 18)d = 1;else if ((a + b) * c == 18)d = 1;else if (c != 0 && (a + b) / c == 18)d = 1;else if (a - b + c == 18)d = 1;else if (a - b - c == 18)d = 1;else if (a - b * c == 18)d = 1;else if (c != 0 && a - b / c == 18)d = 1;else if ((a - b) * c == 18)d = 1;else if (c != 0 && (a - b) / c == 18)d = 1;else if (a * b + c == 18)d = 1;else if (a * b - c == 18)d = 1;else if (a * b * c == 18)d = 1;else if (c != 0 && a * b / c == 18)d = 1;else if (a * (b + c) == 18)d = 1;else if (a * (b - c) == 18)d = 1;else if (a / b + c == 18)d = 1;else if (a / b - c == 18)d = 1;else if (a / b * c == 18)d = 1;else if (c != 0 && b != 0 && a / b / c == 18)d = 1;else if (b + c != 0 && a / (b + c) == 18)d = 1;else if (b - c != 0 && a / (b - c) == 18)d = 1;a = x3; b = x1; c = x2;if (a + b + c == 18)d = 1;else if (a + b - c == 18)d = 1;else if (a + b * c == 18)d = 1;else if (c != 0 && a + b / c == 18)d = 1;else if ((a + b) * c == 18)d = 1;else if (c != 0 && (a + b) / c == 18)d = 1;else if (a - b + c == 18)d = 1;else if (a - b - c == 18)d = 1;else if (a - b * c == 18)d = 1;else if (c != 0 && a - b / c == 18)d = 1;else if ((a - b) * c == 18)d = 1;else if (c != 0 && (a - b) / c == 18)d = 1;else if (a * b + c == 18)d = 1;else if (a * b - c == 18)d = 1;else if (a * b * c == 18)d = 1;else if (c != 0 && a * b / c == 18)d = 1;else if (a * (b + c) == 18)d = 1;else if (a * (b - c) == 18)d = 1;else if (a / b + c == 18)d = 1;else if (a / b - c == 18)d = 1;else if (a / b * c == 18)d = 1;else if (c != 0 && b != 0 && a / b / c == 18)d = 1;else if (b + c != 0 && a / (b + c) == 18)d = 1;else if (b - c != 0 && a / (b - c) == 18)d = 1;a = x3; b = x2; c = x1;if (a + b + c == 18)d = 1;else if (a + b - c == 18)d = 1;else if (a + b * c == 18)d = 1;else if (c != 0 && a + b / c == 18)d = 1;else if ((a + b) * c == 18)d = 1;else if (c != 0 && (a + b) / c == 18)d = 1;else if (a - b + c == 18)d = 1;else if (a - b - c == 18)d = 1;else if (a - b * c == 18)d = 1;else if (c != 0 && a - b / c == 18)d = 1;else if ((a - b) * c == 18)d = 1;else if (c != 0 && (a - b) / c == 18)d = 1;else if (a * b + c == 18)d = 1;else if (a * b - c == 18)d = 1;else if (a * b * c == 18)d = 1;else if (c != 0 && a * b / c == 18)d = 1;else if (a * (b + c) == 18)d = 1;else if (a * (b - c) == 18)d = 1;else if (a / b + c == 18)d = 1;else if (a / b - c == 18)d = 1;else if (a / b * c == 18)d = 1;else if (c != 0 && b != 0 && a / b / c == 18)d = 1;else if (b + c != 0 && a / (b + c) == 18)d = 1;else if (b - c != 0 && a / (b - c) == 18)d = 1;if (d == 1)cout << "Yes";else cout << "No";cout << endl;}return 0; }

這是我的第一種做法，很蠢但至少過了。

#include<iostream> #include<cmath> using namespace std; bool isSame(double a, double b) {if (abs(a - b) <= 1e-7){return true;}else{return false;} } bool can2(double result, int first, int second) {if (first + second == result|| first - second == result|| second - first == result|| first * second == result|| second != 0 && isSame(1.0 * first / second, result)|| first != 0 && isSame(1.0 * second / first, result)){return true;}else{return false;} } bool can31(double result, int first, int second, int third) {if (can2(result - first, second, third)|| can2(first - result, second, third)|| can2(result + first, second, third)|| (result == 0 && first == 0 || first != 0) && can2(result*first, second, third)|| first != 0 && can2(result / first, second, third)|| result != 0 && first != 0 && can2(first / result, second, third)){return true;}else{return false;} } bool can3(double result, int first, int second, int third) {if (can31(result, first, second, third)|| can31(result, second, first, third)|| can31(result, third, first, second)){return true;}else{return false;} } int main() {const double result = 18;int n;cin >> n;while (n--){int a, b, c;cin >> a >> b >> c;if (can3(result, a, b, c)){cout << "Yes" << endl;}else{cout << "No" << endl;}}return 0; }

這是遞歸的做法，比較高難。

第六題：字符陣列

輸出m行m列由字符組成的陣列。其特點是按照字母表順序蛇形走位，第一行是順序，第二行是倒序，…。另外，當用到字符Z后，下一個字符是A。

舉個栗子，就是3的話，會變成
ABC
FED
GHI
這個按坐標算出來在自己所在的字符串第幾個，再輸出ascii對應的字母就行。

#include<iostream> #include<string> #include<cmath> #include<bitset> #include<algorithm> using namespace std;int main() {int f;cin >> f;while (f--){int m;cin >> m;for (int i = 1; i <= m; i++){for (int p = 1; p <= m; p++){if (i % 2 == 1){int x = p + m*(i - 1);char n = (x - 1) % 26 + 65;cout << n;}else{int x = (m - p + 1) + m*(i - 1);char n = (x - 1) % 26 + 65;cout << n;}}cout << endl;}}return 0; }

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