超短题解极致压行

2024-06-16 17:07:06

发布于：广东

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首先，我们先来 正经地 分析一下题目。
要求一个最小的 $n$ ，使得 $S_n > k$ 也就是求 $\sum_{i = 1}^{n}{\frac 1 i} > k$ 中最小的 $n$ 。
根据上述形式化题意，可以知道需要使用循环求出~~这一大坨东西~~这个式子的和，再判断这个和是否大于 $k$ 。如果大于时，输出 $n$ 。根据上述思路，我们可以写出以下代码：

# include <iostream>
using namespace std;
int main(){
    int k,n = 1; cin >> k;
    double sum = 0;
    while (sum <= k) sum += (1.00 / n++);
    cout << --n;
    return 0;
}

再经过极致压行，就可以变成这样：

# include <iostream>
using namespace std; int main(){ int k,n = 1; cin >> k; double sum = 0; while (sum <= k) {sum += (1.00 / n++);} cout << --n; return 0;}

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TC.lt

#include<iostream>
using namespace std;
int main(){
double s;
int a,k; //要用整数类型
s=0;
cin>>k;
while(s<=k){
a=a+1;
s=s+1.0/a;//要加1.0，加1会炸
}
cout<<a<<endl;
}

1周前来自浙江
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2025-10-03 来自广东
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