这道题让我差点怀疑自己高考没考过物理
题意中
he measures the resistance of any two endpoints of it, the resistance values are all $2a$
指的是在三角形中电阻为 $2a$ 而不是边上的电阻为 $2a$
实际上每条边的电阻R为
$\frac{1}{R} + \frac{1}{2R} = 2a$
可以求得$R = 3a$
所以可以得到递推公式
$a{n+1} = \frac{1}{ \frac{1}{ \frac{1}{ \frac{1}{a{n}} + \frac{4}{3}} + 3} + \frac{1}{3}}$
通过python打表
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| res = 5 / 3
print('%.20f' % res)
for i in range(20):
res = 1 / ((1 / (1 / (1 / res + 4 / 3) + 3)) + 1 / 3)
print('%.20f' % res)
|
得到
1.66666666666666674068
1.61904761904761906877
1.61805555555555535818
1.61803444782168193150
1.61803399852180329610
1.61803398895790206957
1.61803398875432269399
1.61803398874998927148
1.61803398874989712297
1.61803398874989490253
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
1.61803398874989468048
这是 $a = 1$ 的情况，最后乘上 a 就行
很明显了，直接打表就行，借助一下字符串流
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| #include <bits/stdc++.h>
using namespace std;
vector<double> res;
void init() {
res.push_back(1.66666666666666674068);
res.push_back(1.61904761904761906877);
res.push_back(1.61805555555555535818);
res.push_back(1.61803444782168193150);
res.push_back(1.61803399852180329610);
res.push_back(1.61803398895790206957);
res.push_back(1.61803398875432269399);
res.push_back(1.61803398874998927148);
res.push_back(1.61803398874989712297);
res.push_back(1.61803398874989468048);
res.push_back(1.61803398874989468048);
res.push_back(1.61803398874989468048);
res.push_back(1.61803398874989468048);
res.push_back(1.61803398874989468048);
res.push_back(1.61803398874989468048);
}
void solve() {
int t;
cin >> t;
init();
while (t--) {
string str;
double a;
cin >> str >> a;
if (str.length() > 2) {
cout << fixed << setprecision(10) << res.back() * a << endl;
continue;
}
stringstream ss(str);
int n;
ss >> n;
if (n > res.size() - 1) {
cout << fixed << setprecision(10) << res.back() * a << endl;
} else {
cout << fixed << setprecision(10) << res[n - 1] * a << endl;
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long long test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}
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