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
",ans+t);
106     }
107     return 0;
108 }