给定一系列正整数,请按要求对数字进行分类,并输出以下5个数字:
- A1 = 能被5整除的数字中所有偶数的和;
- A2 = 将被5除后余1的数字按给出顺序进行交错求和,即计算n1-n2+n3-n4...;
- A3 = 被5除后余2的数字的个数;
- A4 = 被5除后余3的数字的平均数,精确到小数点后1位;
- A5 = 被5除后余4的数字中最大数字。
输入格式:
每个输入包含1个测试用例。每个测试用例先给出一个不超过1000的正整数N,随后给出N个不超过1000的待分类的正整数。数字间以空格分隔。
输出格式:
对给定的N个正整数,按题目要求计算A1~A5并在一行中顺序输出。数字间以空格分隔,但行末不得有多余空格。
若其中某一类数字不存在,则在相应位置输出“N”。输入样例1:
13 1 2 3 4 5 6 7 8 9 10 20 16 18
输出样例1:30 11 2 9.7 9
输入样例2:8 1 2 4 5 6 7 9 16
输出样例2:N 11 2 N 9
-
-
#include "iostream" #include "vector" #include<iomanip> // 设计小数的位数是要包含这个头文件 //#include "stdio.h" using namespace std; int main() { int tmp_input = 0; vector<int> input; vector <int> A1, A2, A3, A4, A5; int sum_A1 = 0; int sum_A2 = 0; int count_A3 = 0; float ave_A4 = 0.0; int max_A5 = 0; int count = 0; int num; cin >> num; while (num--) { cin >> tmp_input; input.push_back(tmp_input); } for (int i = 0; i < input.size(); i++) { switch (input[i] % 5) { case 0: A1.push_back(input[i]); break; case 1: A2.push_back(input[i]); break; case 2: A3.push_back(input[i]); break; case 3: A4.push_back(input[i]); break; case 4: A5.push_back(input[i]); break; default: break; } } for (int i = 0; i < A1.size(); i++) { if (A1[i] % 2 == 0) sum_A1 = sum_A1 + A1[i]; } for (int i = 0; i < A2.size(); i++) { if (i % 2 == 0) sum_A2 = sum_A2 + A2[i]; else sum_A2 = sum_A2 - A2[i]; } for (int i = 0; i < A3.size(); i++) { count_A3++; } float sum_tmp = 0.0; for (int i = 0; i < A4.size(); i++) { sum_tmp += A4[i]; } ave_A4 = sum_tmp / (float)(A4.size()); for (int i = 0; i < A5.size(); i++) { if (max_A5 < A5[i]) { max_A5 = A5[i]; } } if (A1.size() == 0||sum_A1 ==0) cout << "N "; else cout << sum_A1<<" "; if (A2.size() == 0) cout << "N "; else cout << sum_A2 << " "; if (A3.size() == 0) cout << "N "; else cout << count_A3 << " "; if (A4.size() == 0) cout << "N "; else cout << setiosflags(ios::fixed) << setprecision(1) << ave_A4 << " "; // C++实现精度位数的设定 if (A5.size() == 0) cout << "N"; else cout << max_A5; system("pause"); return 0; }