1 条题解
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0
C++ :
#include<bits/stdc++.h> using namespace std; int main() { int n,a[110],s = 0,ma = 0; cin>>n; int i; for(i = 0; i < n; i++) { cin>>a[i]; if(a[i] > a[ma]){ ma = i; } } cout<<fixed<<setprecision(2)<<3.14 * a[ma] * a[ma]; cout<<" "; cout<<ma + 1; }
Java :
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int num1 = scanner.nextInt(); // int num2 = scanner.nextInt(); int[] a = new int[num1]; int x = 0,y = 0; double s = 0; for (int i = 0; i < num1; i++) { a[i] = scanner.nextInt(); } double PI = 3.14; for (int i = 0; i < num1; i++) { if(MAX(a) == a[i]){ s = PI * a[i] * a[i]; System.out.println(String.format("%.2f",s) + " " + (i + 1)); break; } } } public static boolean judge(int i,int j){ int a,s = 0; while(i != 0){ a = i % 10; s = s + a; i /= 10; } if(s == j)return true; else return false; } public static void paiXu(int[] a){ int temp; for(int i = 0;i < a.length - 1;i++){ for(int k = 0;k < a.length - i - 1;k++){ if(a[k] > a[k + 1]){ temp = a[k]; a[k] = a[k + 1]; a[k + 1] = temp; } } } } public static int MAX(int[] a){ int max = a[0]; for (int i = 0; i < a.length; i++) { if(max < a[i]){ max = a[i]; } } return max; } public static int MIN(int[] a){ int min = a[0]; for (int i = 0; i < a.length; i++) { if(min > a[i]){ min = a[i]; } } return min; } }
Python :
n = int(input()) s = input().split() l = [] for v in s: l.append(int(v)) # 遍历l列表,找出最大半径的数,这个数得出的面积一定最大 ma = max(l) p = ma * ma * 3.14 print('%.2f' % p, end=' ') for i in range(0, n): if l[i] == ma: print(i + 1) break
- 1
信息
- ID
- 2682
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者