1 条题解
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0
C :
#include <stdio.h> #include <stdlib.h> /* run this program using the console pauser or add your own getch, system("pause") or input loop */ void main(int argc, char *argv[]) { int a[10]; int i,n,y; double s; s=0; scanf("%d",&n); for(i=0;i<n;i++){ scanf("%d",&a[i]); } for(i=0;i<n;i++){ s=s+a[i]; } for(i=0;i<n;i++){ if(s>100&&a[i]==a[n-1]) printf("%.2f",(s-100)*0.9+100); } }C++ :
#include <iostream> #include <iomanip> #include <cmath> using namespace std; int main(){ int a[100],n,s=0,m,x; cin>>n; for(int i=0;i<n;i++){ cin>>a[i]; s+=a[i]; } cout<<fixed<<setprecision(2)<<100+((s-100)*0.9)<<endl; }Pascal :
var s:real; n,m,i:longint; begin readln(n); for i:=1 to n do begin read(m); s:=s+m; //读入并累加 end; if s>100 then s:=s-(s-100)+(s-100)*0.9; writeln(s:0:2); end.Java :
import java.util.Scanner; public class Main { public static void main(String args[]){ Scanner sc = new Scanner(System.in); int n=sc.nextInt(); int[] a=new int[n]; for(int i=0;i<n;i++){ a[i]=sc.nextInt(); } double s=0.00; for(int i=0;i<n;i++){ s+=a[i]; } if(s<=100){ System.out.println(String.format("%.2f", s)); }else{ System.out.println(String.format("%.2f", (s-100)*0.9+100)); } } }Python :
n=int(input()) l1=list(map(int,input().split())) def qiuHe(n): s=0 for i in l1: s+=i return s s=qiuHe(l1) if s>100: s=(s-100)*0.9+100 print('%.2f'% s) else: print(s)
- 1
信息
- ID
- 155
- 时间
- 1000ms
- 内存
- 512MiB
- 难度
- 10
- 标签
- 递交数
- 3
- 已通过
- 3
- 上传者