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 }
