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[CodeForces 344C Rational Resistance]YY,证明

时间:2015-08-12 18:30:24      阅读:114      评论:0      收藏:0      [点我收藏+]

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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

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