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=50010,M=2000010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=1;
    LL ret=0;
    for(i=a=b=0;1;){
        la=upa-a+1;lb=upb-b+1;
        if(la<lb){
            if(ok && i+la>=late){
                t=ret+(late-i)*abs(a-b);
                ok=0;
                if(!flag)break;
            }
            ret+=(LL)la*abs(a-b);
            a=0;b=(b+la)%B;
        }
        else{
            if(ok && i+lb>=late){
                t=ret+(late-i)*abs(a-b);
                ok=0;
                if(!flag)break;
            }
            ret+=(LL)lb*abs(a-b);
            b=0;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-1,B-1,n/cir)*(n/cir);

        printf("%I64d
",ans+t);
    }
    return 0;
}