POJ--2142[The Balance] 扩展欧几里德

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题意：Ax+By=C;求一组解x,y使abs(x)+abs(y)最小，若abs(x)+abs(y)相等,则使(A*abs(x)+B*abs(y))最小

解法：在有解的情况下先得到一组解使x为最小正数,然后分别向上和向下枚举直到x,y异号停止,中间不停更新答案。每次x-dx则y+dy

CODE：

/*扩展欧几里德*//*AC代码：0ms*/#include <iostream>#include <cstdio>#include <algorithm>#include <memory.h>#include <cmath>using namespace std;int A,B,C;int ans_x,ans_y,Min;int extgcd(int a,int b,int &x,int &y){if(b==0) {x=1;y=0;return a;}int d=extgcd(b,a%b,x,y);int t=x;x=y;y=t-a/b*y;return d;}int main(){int w,dx,dy,X,Y,x,y,temp;while(scanf("%d%d%d",&A,&B,&C)!=EOF){if(A==0&&B==0&&C==0) break;w=extgcd(A,B,X,Y);if(C%w) printf("no solution\n");else{X*=C/w;dx=B/w;dy=A/w;X=(X%dx+dx)%dx;Y=(C-A*X)/B;ans_x=abs(X);ans_y=abs(Y);Min=abs(X)+abs(Y);//printf("@%d %d %d %d\n",X,Y,dx,dy);//从X,Y开始上下枚举,直到x,y异号x=X;y=Y;while(true)//向上枚举{if(x>0&&y<0) break;x+=dx;y-=dy;temp=abs(x)+abs(y);if(temp<Min)//更新答案{ans_x=abs(x);ans_y=abs(y);Min=temp;}else if(temp==Min){if((A*abs(x)+B*abs(y))<(A*ans_x+B*ans_y)){ans_x=abs(x);ans_y=abs(y);}}}x=X;y=Y;while(true)//向下枚举 { if(x<0&&y>0) break;x-=dx;y+=dy;temp=abs(x)+abs(y);if(temp<Min)//更新答案{ans_x=abs(x);ans_y=abs(y);Min=temp;}else if(temp==Min){if((A*abs(x)+B*abs(y))<(A*ans_x+B*ans_y)){ans_x=abs(x);ans_y=abs(y);}}}printf("%d %d\n",ans_x,ans_y);}}return 0;}