HDU-4611 Balls Rearrangement 循环节,模拟
2013-07-28 18:14
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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。。。
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\n",ans+t); } return 0; }
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