HDU-4611 Balls Rearrangement 循环节,模拟

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4611

　　先求出循环节，然后比较A和B的大小模拟过去。。。

 //STATUS:C++_AC_15MS_436KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 #pragma comment(linker,"/STACK:102400000,102400000")
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=,M=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int T;
 LL n,A,B,t,late;

 LL slove(LL upa,LL upb,int flag)
 {
     int i,j,a,b,la,lb,ok=;
     LL ret=;
     for(i=a=b=;;){
         la=upa-a+;lb=upb-b+;
         if(la<lb){
             if(ok && i+la>=late){
                 t=ret+(late-i)*abs(a-b);
                 ok=;
                 if(!flag)break;
             }
             ret+=(LL)la*abs(a-b);
             a=;b=(b+la)%B;
         }
         else{
             if(ok && i+lb>=late){
                 t=ret+(late-i)*abs(a-b);
                 ok=;
                 if(!flag)break;
             }
             ret+=(LL)lb*abs(a-b);
             b=;a=(a+lb)%A;
         }
         i+=Min(la,lb);
         if(a==b)break;
     }
     return ret;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j;
     LL ans,cir;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%I64d%I64d%I64d",&n,&A,&B);

         cir=lcm(A,B);
         late=n-n/cir*cir;
         ans=slove(A-,B-,n/cir)*(n/cir);

         printf("%I64d\n",ans+t);
     }
     return ;
 }