1 条题解
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0
C :
#include<stdio.h> int main(){ int N; while(scanf("%d",&N)!=EOF){ double z=2.0,m=1.0,t; int i; double sum=0.0; for(i=1;i<=N;i++){ sum=sum+z/m; t=z; z=z+m; m=t; } printf("%.2f\n",sum); } return 0; }
C++ :
#include <iostream> #include <iomanip> using namespace std; int main() { int n; double sum; double F1, F2, F; while (cin>>n) { F1 = 1; F2 = 2; sum = F2/F1; for (int i=2; i<=n; i++) { F = F1; F1 = F2; F2 = F1 + F; sum += F2/F1; } cout <<setprecision(2) <<std::fixed<< sum<<endl; } return 0; }
Java :
import java.util.Scanner; public class Main{ public static void main(String[] args){ Scanner input = new Scanner(System.in); while(input.hasNextInt()){ int N = input.nextInt(); int x,a = 1,b = 2; double sum = 0; for(x = 1;x <= N; x++){ sum = sum + ((double)b / a); int y = a; a = b; b = b + y; } String he = String.format("%.2f",sum); System.out.println(he); } } }
- 1
信息
- ID
- 1433
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者