bzoj2712 -- 类欧几里得算法
与bzoj2187类似，不过是要先将小数转化成四舍五入前的分数
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<cmath> 5 using namespace std; 6 #define ll long long 7 struct Node{ 8 ll x,y; 9 Node(){} 10 Node(ll x,ll y):x(x),y(y){} 11 }a,b,c; 12 int n; 13 ll g,i,j,k,m,x,p; 14 inline ll _Min(ll x,ll y){return x<y?x:y;} 15 inline ll Gcd(ll x,ll y){return y==0?x:Gcd(y,x%y);} 16 inline ll F(ll x,ll y){return (x-1)/y+1;} 17 inline Node Solve(Node a,Node b){ 18 g=Gcd(a.x,a.y);a.x/=g;a.y/=g; 19 g=Gcd(b.x,b.y);b.x/=g;b.y/=g; 20 if(a.x==0)return Node(1,b.y/b.x+1); 21 if(a.x/a.y+2<=F(b.x,b.y))return Node(a.x/a.y+1,1); 22 if(a.x/a.y>=1){ 23 int x=a.x/a.y; 24 Node c=Solve(Node(a.x%a.y,a.y),Node(b.x-x*b.y,b.y)); 25 return Node(c.x+x*c.y,c.y); 26 } 27 if(b.x>b.y)return Node(1,1); 28 Node c=Solve(Node(b.y,b.x),Node(a.y,a.x)); 29 return Node(c.y,c.x); 30 } 31 int main() 32 { 33 while(scanf("%d 0.%lld",&n,&x)!=EOF){ 34 if(x==0){printf("1\n");continue;} 35 for(p=1;n>=0;n--)p*=10; 36 a=Node(x*10-5,p);b=Node(x*10+5,p); 37 g=Gcd(a.x,a.y);a.x/=g;a.y/=g; 38 g=Gcd(b.x,b.y);b.x/=g;b.y/=g; 39 c=Solve(a,b); 40 printf("%lld\n",_Min(c.y,a.y)); 41 } 42 return 0; 43 }
