题解

用的正常方法

#include<cstdio> #include<iostream> #include<cmath> using namespace std; int main(){ int a,b=0,e=114514,f=0; for(int i=1;i<=4;i++){ cin>>a; if(a>f)f=a; if(a<e)e=a; b+=a; } printf("%.2f",(b-f-e)*1.0/2); return 0; }

#include <bits/stdc++.h> using namespace std; int a,b,c,d,sum; int main() { cin>>a>>b>>c>>d; sum=a+b+c+d; int maxx,minn; maxx=max(max(max(a,b),c),d); minn=min(min(min(a,b),c),d); printf("%.2f",(sum-minn-maxx)/(2.00)); return 0; }

//#include<iostream> #include<cstdio> #include<algorithm> //#include<cmath> //#include<string.h> using namespace std; int main() { int a,b,c,d; scanf("%d%d%d%d",&a,&b,&c,&d); double sum=a+b+c+d; sum=sum-max(max(a,b),max(c,d))-min(min(a,b),min(c,d)); printf("%.2f",sum/2); return 0; }

#include<iostream> using namespace std; int main(){ double a,b,c,d,e,f,g,h,i,j,l; cin>>a>>b>>c>>d; e = max(a,b); f = max(c,d); g = max(e,f); h = min(a,b); i = min(c,d); j = min(h,i); if(ag){ a=0; }else if(bg){ b=0; }else if(cg){ c=0; }else if(dg){ d=0; } if(aj){ a=0; }else if(bj){ b=0; }else if(cj){ c=0; }else if(dj){ d=0; } l = (a+b+c+d)/2; printf("%.2lf",l); return 0; } 注意这里要用double。

#include<iostream> using namespace std; int main(){ int a,d=0,x=10000; float n=0; for(int i=0;i<4;i++){ cin>>a; n+=a; if(a>d) {d=a;} if(a<x) {x=a;} } n=n-d-x; printf("%.2f",n/2*1.0); return 0; }