2 条题解

  • 2
    @ 2025-12-5 10:38:20
    #include<bits/stdc++.h>
    using namespace std;
    int main(){
        int a[5]={0},n,c;
    	cin>>n;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>c;
    		if(c<=18)        a[1]++;
    		if(19<=c&&c<=35) a[2]++;
    		if(36<=c&&c<=60) a[3]++;
    		if(c>60)         a[4]++;
    	}
    	printf("%.2lf%%\n",double(a[1])/n*100);
    	printf("%.2lf%%\n",double(a[2])/n*100);
    	printf("%.2lf%%\n",double(a[3])/n*100);
    	printf("%.2lf%%\n",double(a[4])/n*100);
     	return 0;
    }
    
    • -2
      @ 2025-12-4 20:43:37

      #include #include using namespace std;

      int main() { int N; cin >> N;

      int age;
      int count1 = 0, count2 = 0, count3 = 0, count4 = 0;  // 分别对应四个年龄段
      
      for (int i = 0; i < N; i++) {
          cin >> age;
          if (age >= 0 && age <= 18) {
              count1++;
          } else if (age >= 19 && age <= 35) {
              count2++;
          } else if (age >= 36 && age <= 60) {
              count3++;
          } else {
              count4++;
          }
      }
      
      cout << fixed << setprecision(2);
      cout << (count1 * 100.0 / N) << "%" << endl;
      cout << (count2 * 100.0 / N) << "%" << endl;
      cout << (count3 * 100.0 / N) << "%" << endl;
      cout << (count4 * 100.0 / N) << "%" << endl;
      
      return 0;
      

      }

      • 1

      信息

      ID
      6947
      时间
      1000ms
      内存
      128MiB
      难度
      1
      标签
      递交数
      8
      已通过
      6
      上传者