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。。。

  1 //STATUS:C++_AC_15MS_436KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 #pragma comment(linker,"/STACK:102400000,102400000")

 24 using namespace std;

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef __int64 LL;

 34 typedef unsigned __int64 ULL;

 35 //const

 36 const int N=50010,M=2000010;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=100000,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-8;

 41 const double OO=1e15;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 int T;

 59 LL n,A,B,t,late;

 60 

 61 LL slove(LL upa,LL upb,int flag)

 62 {

 63     int i,j,a,b,la,lb,ok=1;

 64     LL ret=0;

 65     for(i=a=b=0;1;){

 66         la=upa-a+1;lb=upb-b+1;

 67         if(la<lb){

 68             if(ok && i+la>=late){

 69                 t=ret+(late-i)*abs(a-b);

 70                 ok=0;

 71                 if(!flag)break;

 72             }

 73             ret+=(LL)la*abs(a-b);

 74             a=0;b=(b+la)%B;

 75         }

 76         else{

 77             if(ok && i+lb>=late){

 78                 t=ret+(late-i)*abs(a-b);

 79                 ok=0;

 80                 if(!flag)break;

 81             }

 82             ret+=(LL)lb*abs(a-b);

 83             b=0;a=(a+lb)%A;

 84         }

 85         i+=Min(la,lb);

 86         if(a==b)break;

 87     }

 88     return ret;

 89 }

 90 

 91 int main()

 92 {

 93  //   freopen("in.txt","r",stdin);

 94     int i,j;

 95     LL ans,cir;

 96     scanf("%d",&T);

 97     while(T--)

 98     {

 99         scanf("%I64d%I64d%I64d",&n,&A,&B);

100 

101         cir=lcm(A,B);

102         late=n-n/cir*cir;

103         ans=slove(A-1,B-1,n/cir)*(n/cir);

104 

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

106     }

107     return 0;

108 }