[扩展欧拉函数] POJ2142The Balance
程序员文章站
2022-03-27 21:39:09
题目有两种砝码a,b,要用ab称出c的质量,计算ab的最少个数题目链接:http://poj.org/problem?id=2142思路还没学LaTeX代码#include#include#include#include#include#include#include#includ...
题目
有两种砝码a,b,要用ab称出c的质量,计算ab的最少个数
题目链接:http://poj.org/problem?id=2142
思路
还没学LaTeX
代码
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<ctime>
#include<iostream>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<iomanip>
#include<list>
#include<bitset>
#include<sstream>
#include<fstream>
#include<complex>
#include<algorithm>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
int exgcd(int a,int b,int &x,int &y){
if(!b){
x=1;
y=0;
return a;
}
int g=exgcd(b,a%b,x,y);
int temp=x;
x=y;
y=temp-(a/b)*y;
return g;
}
int main(){
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int a,b,c;
while(cin>>a>>b>>c){
if(a==0&&b==0&&c==0) break;
bool flag=0;
if(a<b){
swap(a,b);
flag=1;
}
int x,y;
int gcd=exgcd(a,b,x,y);
c/=gcd,a/=gcd,b/=gcd;
y*=c,x*=c;
int t=y/a;
int minn=INF;
int res1=0,res2=0;
for(int i=t-1;i<=t+1;i++){
int xx=abs(x+i*b);
int yy=abs(y-i*a);
//cout<<xx<<" "<<yy<<endl;
if(xx+yy<minn){
minn=xx+yy;
res1=xx,res2=yy;
}
}
if(flag) cout<<res2<<" "<<res1<<endl;
else cout<<res1<<" "<<res2<<endl;
}
return 0;
}
本文地址:https://blog.csdn.net/kosf_/article/details/107419101
上一篇: 5+ API,APP更新逻辑部分