十三行过
var
i,j,b,n:integer;
a:array[1..10]of integer;
begin
for i:=1 to 10 do
read(a[i]);
readln(n);
n:=n+30;
j:=0;
for i:=1 to 10 do
if a[i]<=n then inc(j);
writeln(j);
end.

C语言 题解 已通过
#include<cstdio>
int a[10];
int main()
{
int x,b,n=0,c=0;
while(scanf("%d",&x)==1)
{char ch;
ch=getchar();
a[n++]=x;
if (ch=='\n')
{
break;
}
}
scanf("%d",&b) ;
for(int i=0;i<n;i++)
{
if((a[i]<=(b+30)))
{
c=c+1;
}
else
{
c=c+0;

}
}
printf("%d",c);
return 0;
}

#py 的第一条题解（或者我没注意）
s=input()
a,b,c,d,e,f,g,h,i,j=s.split(' ')
k=input()
k=int(k)
ans = 0
if int(a)<=k+30:
ans=ans+1
if int(b)<=k+30:
ans=ans+1
if int(c)<=k+30:
ans=ans+1
if int(d)<=k+30:
ans=ans+1
if int(e)<=k+30:
ans=ans+1
if int(f)<=k+30:
ans=ans+1
if int(g)<=k+30:
ans=ans+1
if int(h)<=k+30:
ans=ans+1
if int(i)<=k+30:
ans=ans+1
if int(j)<=k+30:
ans=ans+1
print(ans)

var i,x,ans:longint; a:array[1..10] of longint;
begin
for i:=1 to 10 do
read(a[i]);
readln(x);
for i:=1 to 10 do
if (x+30)>=a[i] then inc(ans);
writeln(ans);
end.

#include<stdio.h>
#include<stdlib.h>
int main()
{
int a,b[15],ans=0,i;
for(i=1;i<=10;i++)
{
scanf("%d",&b[i]);
}
scanf("%d",&a);
a=a+30;
for(i=1;i<=10;i++)
{
if(a>=b[i])
{
ans++;
}
}
printf("%d",ans);
}

#include<stdio.h>
#include<stdlib.h>
int main()
{
int a,b[15],ans=0,i;
for(i=1;i<=10;i++)
{
scanf("%d",&b[i]);
}
scanf("%d",&a);
a=a+30;
for(i=1;i<=10;i++)
{
if(a>=b[i])
{
ans++;
}
}
printf("%d",ans);
}

#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
int main ()
{
priority_queue<int,vector<int>,greater<int> > pq;
int n,x,sum=0;
for(int i=0;i<10;i++)
{
cin>>x;
pq.push(x);
}
cin>>x;
while(pq.size()>1)
{
if(x+30>=pq.top())
sum++;
else
break；
pq.pop();
}
cout<<sum;
return 0;
}

算是比较简单的题了吧，用一个数组就能过了。

#include <iostream>
using namespace std ;
int main ()
{
    int total=0;
    int a[10]={0};      
    int max;
          cin>>a[0]>>a[1]>>a[2]>>a[3]>>a[4]>>a[5]>>a[6]>>a[7]>>a[8]>>a[9];
    
    cin>>max;
    for (int n=0;n<10;n++)
    {
        if (a[n]<=max+30)
        total++;
    }
    cout<<total;
    return 0;
}

不难
#include<cstdio>
#include<iostream>
#include <algorithm>

using namespace std;
int apple[15];
int main(){
for(int i = 1;i<11;i++){
cin >>apple[i];
}
int taotao;
cin >> taotao;
int counter = 0;
for(int i = 1;i < 11;i++){
if(taotao + 30 >= apple[i]){
counter++;
}
}
cout << counter << endl;
return 0;
}

不用那么麻烦，直接用10个变量比较就行了，又简单又容易理解。
#include<iostream>
using namespace std;
int main()
{
int n,l=1,o,p,q,w,v,z=0,x,m,s,t,r,y,u,k;
int a,b,c,d,e,f,g,h,i,j;
cin>>a>>b>>c>>d>>e>>f>>g>>h>>i>>j;
cin>>n;
n+=30;
while(l<=10)
{
if(a<=n){z++;a+=n;}
if(b<=n){z++;b+=n;}
if(c<=n){z++;c+=n;}
if(d<=n){z++;d+=n;}
if(e<=n){z++;e+=n;}
if(f<=n){z++;f+=n;}
if(g<=n){z++;g+=n;}
if(h<=n){z++;h+=n;}
if(i<=n){z++;i+=n;}
if(j<=n){z++;j+=n;}
l++;
}
cout<<z;
return 0;
}

Python大法好

count = 0
n = raw_input()
list = n.split(' ')
t = raw_input()
for nn in list:
    if int(t) + 30 >= int(nn):
        count += 1;
print count

AC！！！GG!!!没毛病
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int s;
int l=0;
int a[11];
int i;
for(int i=1;i<=10;i++)
cin>>a[i];
cin>>s;
for(int i=1;i<=10;i++){
if(a[i]<=(s+30))
l++;}
cout<<l;
return 0;
}

#include<iostream>
using namespace std;
int a[100];
int main()
{
int i,j,k=0,n;
for(i=0;i<10;i++)
cin>>a[i];
cin>>n;
for(i=0;i<10;i++)
{
if(a[i]<=(n+30))
k++;
}
cout<<k;
return 0;
}

#include<iostream>
using namespace std;
int a[100];
int main()
{
int i,j,k=0,n;
for(i=0;i<10;i++)
cin>>a[i];
cin>>n;
for(i=0;i<10;i++)
{
if(a[i]<=(n+30))
k++;#include<iostream>
using namespace std;
int a[100];
int main()
{
int i,j,k=0,n;
for(i=0;i<10;i++)
cin>>a[i];
cin>>n;
for(i=0;i<10;i++)
{
if(a[i]<=(n+30))
k++;
}
cout<<k;
return 0;
}
}
cout<<k;
return 0;
}

哈哈 这尼玛 这是我除了a+b第一道Ac的题 加油啊

c++短码,113bytes,欢迎挑战更短= =
c++
#include<iostream>
int a[11],s,i=11;int main(){while(i)std::cin>>a[--i];while(i<10)s+=a[++i]<*a+31;std::cout<<s;}


public class TestTT {
public static void main(String args[]){
int[] a=new int[10];
int n=0;

for(int i=0,j=110;i<a.length;i++,j+=5){
a[i]=j;
int b=100;
if(a[i]>b){
b+=30;
}
if(a[i]<=b){
n++;
}

}
System.out.println(n);
}

}

思路：别考虑TT要不要板凳，直接上板凳大法就行了。
#include <iostream>
using namespace std;
int main()
{
int i,a[11],taotao,sum=0;
for(i=1;i<=10;i++)cin>>a[i];
cin>>taotao;
for(i=1;i<=10;i++)sum+=(taotao+30>=a[i]);
cout<<sum;return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;
int n;
int a[10000];
int main(){
int count=0;
for(int i=1;i<=10;i++){
scanf("%d",&a[i]);
}
scanf("%d",&n);
for(int i=1;i<=10;i++){
if(a[i]<=n+30){
count++;
}
}
cout<<count<<endl;
return 0;
}

PROGRAM P1102;
  VAR HIGHAPPLE:ARRAY[1..10] OF INTEGER;
      HEIGHTP:INTEGER;
      I,NUM:INTEGER;
BEGIN
  NUM:=0;
  FOR i:=1 TO 10 DO READ(HIGHAPPLE[I]);
  READLN;
  READ(HEIGHTP);
  FOR I:=1 TO 10 DO HIGHAPPLE[I]:=HIGHAPPLE[I]-30;
  FOR I:=1 TO 10 DO
    BEGIN
      IF HIGHAPPLE[I]<=HEIGHTP THEN INC(NUM);
    END;
  WRITE(NUM);
END.