poj2142(THE BALANCE)（扩展欧几里得入门题）

来源：互联网发布：奥杜尔档案馆数据圆盘编辑：程序博客网时间：2024/06/04 18:39

总共要考虑一下三种情况
1. a *x=b*y+d;
2. a*x=b*y-d;
3. a*x+b*y=d;（其中x和y均为非负整数，可由扩展欧几里得算法求得）

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;#define INF 0xffffffftypedef long long ll;int gcd_ex(int a,int b,int &x,int &y){    if(!b)  {x=1;y=0;return a;}    int dd=gcd_ex(b,a%b,y,x);    y-=a/b*x;    return dd;}int main(int argc, char const *argv[]){    int a,b,d;    while(scanf("%d %d %d",&a,&b,&d)!=EOF&&(a||b||d))    {        int x,y,dd,x1,y1,x2,y2;        dd=gcd_ex(a,b,x,y);        a/=dd,b/=dd,d/=dd;        x1=d*x,y1=-d*y;        while(x1>0) x1-=b,y1-=a;        while(x1<0||y1<0) x1+=b,y1+=a;        x2=d*x,y2=d*y;        if(a>b) {while(y2>=0) x2+=b,y2-=a;            while(y2<0) x2-=b,y2+=a;            if(x2<0) x2=y2=INF;        }        else {while(x2>=0) x2-=b,y2+=a;            while(x2<0) x2+=b,y2-=a;            if(y2<0) x2=y2=INF;        }        x*=-d;y*=d;        while(x>0) x-=b,y-=a;        while(x<0||y<0) x+=b ,y+=a ;        if(x+y<x1+y1&&x+y<x2+y2) printf("%d %d\n",x,y );        else if(x+y>x2+y2&&x1+y1>x2+y2) printf("%d %d\n",x2,y2);        else if(x+y>x1+y1&&x2+y2>x1+y1) printf("%d %d\n",x1,y1 );        else if(x2==INF) {            if(a*x+b*y<a*x1+b*y1) printf("%d %d\n",x,y );        else printf("%d %d\n",x1,y1 );}        else if(x+y<=x1+y1&&x+y<=x2+y2&&a*x+b*y<=a*x1+b*y1&&a*x+b*y<=a*x2+b*y2)         printf("%d %d\n",x,y );        else if(x+y>=x2+y2&&x1+y1>=x2+y2&&a*x+b*y>=a*x2+b*y2&&a*x1+a*y1>=a*x2+b*y2)          printf("%d %d\n",x2,y2);        else if(x+y>=x1+y1&&x2+y2>=x1+y1&&a*x+b*y>=a*x1+b*y1&&a*x2+b*y2>=a*x1+b*y1)         printf("%d %d\n",x1,y1 );    }    return 0;}

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