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题意:给若干个阻值为1的电阻,要得到阻值为a/b的电阻最少需要多少个。
思路:令a=mb+n,则a/b=m+n/b=m+1/(b/n),令f(a,b)表示得到a/b的电阻的答案,由f(a,b)=f(b,a),有:
f(a,b)=a/b + f(a%b,b)=a/b+f(b,a%b)
(1)由于将所有的电阻之间的关系改变一下,串联变成并联,并联变成串联,阻值变成之前的倒数,所以f(a,b)=f(b,a)成立。
(2)现在再证一下:串联变成并联,并联变成串联,阻值变成之前的倒数。考虑任一个电路,一定可以看成两个电路的并联或者两个电路的串联(题目保证了),先假设子问题成立,考虑原来的电路A和电路B,如果A和B串联,则R0=RA+RB,变成并联后R=1/(1/(1/RA)+1/(1/RB))=1/(RA+RB)=1/R0,如果A和B并联,则R=RA*RB/(RA+RB),变成串联后R=1/RA+1/RB=(RA+RB)/(RA*RB)=1/R0,由数学归纳法,所以结论成立
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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ ll f(ll a, ll b) { return b? a / b + f(b, a % b) : 0; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE ll a, b; while (cin >> a >> b) { cout << f(a, b) << endl; } return 0; } |
[CodeForces 344C Rational Resistance]YY,证明
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原文地址:http://www.cnblogs.com/jklongint/p/4724883.html