Codeforces 344C - Rational Resistance
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刚开始有一个阻值为1的电阻。
每次可以将一个电阻和某个阻值为1的电阻串联或者并联来获得一个新的电阻,问至少需要多少个阻值为1的电阻能凑出阻值为
假设当前手头上有
具体见代码
#include<bits/stdc++.h>using namespace std;#define LL long long int main(){ LL a,b; cin>>a>>b; LL tim = 0; do{ if(a > b) swap(a,b); if(a == 1){ tim += b,b = 1; } else{ tim += b / a; b %= a; } } while(a != 1 || b != 1); cout<<tim<<endl; return 0;}
0 0
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