您的位置:首页 > 其它

HDU-4611 Balls Rearrangement 循环节,模拟

2013-07-28 18:14 351 查看
  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\n",ans+t);
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: