1 条题解
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0
C :
#include<stdio.h> int main(){ int n; int i; double sum=100.0,high=100.0; scanf("%d",&n); for(i=1;i<n;i++){ high=high/2; sum+=high*2; } high=high/2; printf("%.4lf %.4lf\n",sum,high); return 0; }
C++ :
#include <stdio.h> int main() { int i, n; float h, tot; h = 100; tot = 0; scanf("%d", &n); /* 模拟前n-1次反弹的过程 */ for (i = 1;i < n;i++) { tot += h; h = h / 2; tot += h; } /* 模拟第n次反弹的过程 */ tot += h; h = h / 2; printf("%.4f %.4f\n", tot, h); return 0; }
Pascal :
var i,n:longint; sum,s:double; begin readln(n); sum:=0; s:=100; for i:=1 to n do begin sum:=sum+s; s:=s/2; sum:=sum+s; end; writeln(sum-s:0:4,' ',s:0:4); end.
Java :
import java.util.*; public class Main { public static void main(String args[]) { Scanner cin=new Scanner(System.in); int n=cin.nextInt(); double[] len=new double[100]; len[0]=100; for(int i=1;i<=n;i++) len[i]=len[i-1]/2.0; double sum=100; for(int i=2;i<=n;i++){ sum+=len[i-2]; } System.out.printf("%.4f %.4f\n",sum,len[n]); } }
Python :
n =input() s = 100 a = 0 for i in range(1, n + 1): s += a a = 100 / 2 ** (i - 1.0) print "%.4f %.4f" % (s, a / 2)
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信息
- ID
- 1282
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者