用高精度就对啦。。。
#include <stdio.h>
int main()
{
long int n=0,k;double s=0;
scanf("%ld",&k);
while(s<=k)
{
n++;
s+=1/(float)n;
}
printf("%ld",n);
return 0;
}

#include "iostream"
#include "cstdio"
using namespace std;
double res = 0;
int k,cnt;
int main()
{
    scanf("%d",&k);
    while(res <= k) cnt++,res+=1.0/cnt;
    cout<<cnt<<endl;
    return 0;
 } 

#include<iostream>
using namespace std;
//int main()
//{
//  int k;
//  cin>>k;
//  unsigned long long n=0;
//  double s=0;
//  do{
//      s+=1.0/++n;
//  }while(s<=k);
//  cout<<n;
//  return 0;
//} 
//1 2 3  4  5  6   7   8    9    10    11    12    13     14     15
//2 4 11 31 83 227 616 1674 4550 12367 33617 91380 248397 675214 1835421
int main()
{
    int n,a[16]={0,2,4,11,31,83,227,616,1674,4550,12367,33617,91380,248397,675214,1835421};
    cin>>n;
    cout<<a[n];
    return 0;
}

 var i,j,k:integer;
      sum:real;
      n:longint;
  begin
     read(k);n:=0; sum:=0;
     while (sum<=k) do
      begin
        n:=n+1;
        sum:=sum+1/n;
      end;
     write(n);
  end.

pacal 一遍AC
难度为3的水题2333
是数据太弱了吧

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<fstream>
using namespace std;
double k,n,i=1;
int main()
{
//freopen("sum.in","r",stdin);
//freopen("sum.out","w",stdout);
cin>>k;
while(n<=k)
{
n+=1/i;
i++;
}
i--;
cout<<i;
return 0;
}

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<fstream>
using namespace std;
double k,n,i=1;
int main()
{
//freopen("sum.in","r",stdin);
//freopen("sum.out","w",stdout);
cin>>k;
while(n<=k)
{
n+=1/i;
i++;
}
i--;
cout<<i;
return 0;
}

#include <iostream>
using namespace std;
int main(){
int num=0;
double sum=0.0,ans=1.0;
cin>>num;
while(sum<=num)
{
sum=sum+(1/ans);
ans++;
}
cout<<ans-1;
}

补上一年前的题解
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
double sn,n,k;
int main()
{
cin>>k;
n=0;sn=0;
while(sn<k)
{
n++;
sn=sn+1/n;
}
cout<<n;
return 0;
}
顶楼上是我一年前的感慨，这代码也是一年前的

#include<iostream>
#include<string>

using namespace std;

int main()
{
int k;
double s=0.0,i=1.0;
cin >> k;
while(1)
{
s += 1.0 / i;
if (s > k)
{
cout << i << endl;
break;
}
else
i++;
}
return 0;
}

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <cstdlib>
using namespace std;
int k,n;
double sn=0.00;
int main() {
cin>>k;
sn=0.00;
for(n=1; sn<=k; n++) {
sn+=(double)(1/(double)n);
//cout<<sn<<endl;
}
cout<<n-1<<endl;
return 0;
}

import re
import sys

def readln():
    return map(int, input().split())

def read():
    s = input()
    k = []
    r = []
    k = re.split(' ',s)
    for i in k:
        r.append(int(i))
    return r

a = int(input())
stp = 0
i = 0
while i<=a:
    stp+=1
    i+=1/stp

print(stp)
        

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
int main(){
    double all=0,i;
    int k;
    scanf("%d",&k);
    for(i=1.0;all<=k;i++)
    {
        all=all+1.0/i;
    }
    printf("%.0lf",i-1);
    return 0;
}

真的，最水的一道noip，n开始的时候开小了，觉得1000就行了，
然后5题就AC2个
加3个零ac了33个，就一直往上加

#include <iostream>
using namespace std;
int main()
{
int k;
double s=0.0,n;
cin>>k;
for(n=1;n<=10000000;n++)
{
s=s+1.0/n;
if(s>k){
cout<<n;
break;}
}
}

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
using namespace std;
int main()
{
int n;
double Sn,k;
scanf("%lf",&k);
n=1;
Sn=0;
while(Sn<=k)
{
Sn+=1.0/n;
n++;
}
printf("%d",n-1);
return 0;
}

水题——级数求和
实际上，考虑到这一题的k值如此之小，我们可以通过直接穷举的方法来得出结果：
穷举
{
Sn=
判断Sn是否***大于***k（是：输出；不是：继续穷举）
}
大家喜闻乐见的代码：

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int k,i;
    double sum; 
    cin>>k;
    sum=0;
    for (i=1;i<=1000000;i++)
    {
        sum=sum+(1.0/i);
        if (sum>k) 
        {
            cout<<i;
            break;
        }
    }
    return 0;
} 

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
int main()
{
int a,b,n=0,k;
double sum=0.0,i=0.0;
scanf("%d",&k);
while(sum<=k)
{
i=i+1.0;
sum=sum+1.0/i;
n++;
}
printf("%d",n);
}

C++:

#include <cstdio>
#include <iostream>

using namespace std;

int main(){
    int k;
    scanf("%d", &k);
    double sum = 0, div = 1;
    while (sum <= k){
        sum += 1.0 / div;
        div++;
    }
    cout << (int)div - 1;
    return 0;
}

语言题？懵.jpg

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <cstdlib>
using namespace std;

int k;
double ans;

int main()
{
    cin>>k;
    int i=0;
    while(ans<=k)
    {
        i++;
        ans+=1.0/double(i);
    }
    cout<<i<<endl;
    return 0;
}
     

无可奈何
这只是正常做法，前面有大牛用欧拉公式==
~~~c++
#include<iostream>
using namespace std;
double k,n,s;
int main(){
cin>>k;
while(s<=k){
s+=1.0/++n;
}
cout<<n;
}
~~~

pascal代码：
var
s,n,k:real;
begin
readln(k);
n:=1;
repeat
s:=s+1/n;
n:=n+1;
until s>k;
writeln(n-1:0:0);
end.