題目鏈接

https://pintia.cn/problem-sets/994805260223102976/problems/994805311146147840

題解

就比較簡單，判斷每個數字是哪種情況，然后進行相應的計算即可。

下面的代碼中其實數組是不必要的，每取一個數字就可以直接進行相應計算。

// PAT BasicLevel 1012 // https://pintia.cn/problem-sets/994805260223102976/problems/994805311146147840#include <iostream> using namespace std;int main() {// 數字個數int n;cin >> n;// 獲取數字int* numArr=new int[n];for(int i=0;i<n;++i){cin >> numArr[i];}// 遍歷數組，計算A1至A5int a1=0,a1Count=0;int a2=0,flag=1,a2Count=0;int a3=0,a3Count=0;double a4Sum=0,a4Count=0;int a5=0,a5Count=0;for(int i=0;i<n;++i){switch(numArr[i]%5){case 0:if(numArr[i]%2==0){a1+=numArr[i];a1Count++;}break;case 1:a2+=flag*numArr[i];flag=-flag;a2Count++;break;case 2:a3++;a3Count++;break;case 3:a4Sum+=numArr[i];a4Count++;break;case 4:if(numArr[i]>a5){a5=numArr[i];a5Count++;}break;}}// 輸出A1至A5if(a1Count>0){cout << a1 << ' ';}else{cout <<"N ";}if(a2Count>0){cout << a2 << ' ';}else{cout << "N ";}if(a3Count>0){cout << a3 << ' ';}else{cout << "N ";}if(a4Count>0){printf("%.1lf ", a4Sum / a4Count);}else{cout << "N ";}if (a5Count > 0){cout << a5;}else{cout << 'N';}delete[] numArr;//system("pause");return 0; }

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作者：@臭咸魚

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轉載于:https://www.cnblogs.com/chouxianyu/p/11300602.html

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