HDU 6351暴力枚举 6354计算几何

Beautiful Now

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1876    Accepted Submission(s): 707

Problem Description

Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m−i. In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?

 

Input

The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.

 

Output

For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.

 

Sample Input

5

12 1

213 2

998244353 1

998244353 2

998244353 3

 

Sample Output

12 21

123 321

298944353 998544323

238944359 998544332

233944859 998544332

 

解析 暴力枚举交换1-min(len-1,k) 次的下标排列情况的   算出每种情况的值   感觉有点超时啊  9!*T*10  每种情况都要除10下分解掉 明明是3e8啊 2.5s很悬啊

 

AC代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n");
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl;
using namespace std;
typedef long long ll;
const ll maxn=1e7+,inf=1e18;
const ll mod=1e9+;
vector<int> per[][];
int vis[],a[],b[];
void init()
{
    for(int i=;i<=;i++)
    {
        for(int j=;j<=i;j++)
            a[j]=j;
        do
        {
            fillchar(vis,);
            int now=,temp=;
            for(int j=;j<=i;j++)
                now=now*+a[j];
            for(int j=;j<=i;j++)
            {
                if(vis[j]==)
                {
                    vis[j]=;
                    for(int k=a[j];k!=j;k=a[k])
                        vis[k]=,temp++;
                }
            }
            per[i][temp].pb(now);
        }while(next_permutation(a+,a+i+));
    }
}
char s[];
int main()
{
    int n,m,t,k,maxx,minn;
    init();
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        minn=maxx=n;
        int len=sprintf(s,"%d",n);
        if(len==)
        {
            printf("%d %d\n",minn,maxx);
            continue;
        }
        reverse(s,s+len);
        if(len<=k)k=len-;
        for(int i=;i<=k;i++)
        {
            for(int j=;j<per[len][i].size();j++)
            {
                int temp=per[len][i][j],cnt=,sum=;
                while(temp)
                {
                   sum=sum*+int(s[temp%-]-'');
                   temp/=;
                }
                //cout<<per[len][i][j]<<" "<<sum<<endl;
                if((int)pow(,len-)>sum)continue;
                maxx=max(maxx,sum);
                minn=min(minn,sum);
            }
        }
        printf("%d %d\n",minn,maxx);
    }
}

http://acm.hdu.edu.cn/showproblem.php?pid=6354

解析 内切的直接把周长加起来  相交的减去原来的弧长 加上另一个圆的弧长  根据余弦定理可以算出来角度  角度*半径就是弧长了

 #include <bits/stdc++.h>
 #define LL long long
 #define PI 3.1415926535897932384626
 #define maxn 1000
 #define EXIT exit(0);
 #define DEBUG puts("Here is a BUG");
 #define CLEAR(name, init) memset(name, init, sizeof(name))
 const double eps = 1e-;
 const int MAXN = (int)1e9 + ;
 using namespace std;
 #define Vector Point
 int dcmp(double x)
 {
     return fabs(x) < eps ?  : (x <  ? - : );
 }
 struct Point
 {
     double x, y;

     Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数
     Point(double x = 0.0, double y = 0.0): x(x), y(y) { }   //构造函数
     friend istream& operator >> (istream& in, Point& P)
     {
         return in >> P.x >> P.y;
     }
     friend ostream& operator << (ostream& out, const Point& P)
     {
         return out << P.x << ' ' << P.y;
     }
     friend Vector operator + (const Vector& A, const Vector& B)
     {
         return Vector(A.x+B.x, A.y+B.y);
     }
     friend Vector operator - (const Point& A, const Point& B)
     {
         return Vector(A.x-B.x, A.y-B.y);
     }
     friend Vector operator * (const Vector& A, const double& p)
     {
         return Vector(A.x*p, A.y*p);
     }
     friend Vector operator / (const Vector& A, const double& p)
     {
         return Vector(A.x/p, A.y/p);
     }
     friend bool operator == (const Point& A, const Point& B)
     {
         return dcmp(A.x-B.x) ==  && dcmp(A.y-B.y) == ;
     }
     friend bool operator < (const Point& A, const Point& B)
     {
         return A.x < B.x || (A.x == B.x && A.y < B.y);
     }
     void in(void)
     {
         scanf("%lf%lf", &x, &y);
     }
     void out(void)
     {
         printf("%lf %lf", x, y);
     }
 };
 template <class T> T sqr(T x)
 {
     return x * x;
 }
 double Dot(const Vector& A, const Vector& B)
 {
     return A.x*B.x + A.y*B.y;    //点积
 }
 double Length(const Vector& A)
 {
     return sqrt(Dot(A, A));
 }
 double Angle(const Vector& A, const Vector& B)
 {
     return acos(Dot(A, B)/Length(A)/Length(B));    //向量夹角
 }
 double Cross(const Vector& A, const Vector& B)
 {
     return A.x*B.y - A.y*B.x;    //叉积
 }
 double Area(const Point& A, const Point& B, const Point& C)
 {
     return fabs(Cross(B-A, C-A));
 }
 Vector normal(Vector x)
 {
     return Point(-x.y, x.x) / Length(x);
 }
 double angle(Vector x)
 {
     return atan2(x.y, x.x);
 }
 Vector vecunit(Vector x)
 {
     return x / Length(x);   //单位向量
 }
 struct Circle
 {
     Point c;    //圆心
     double r;   //半径
     Circle() { }
     Circle(const Circle& rhs): c(rhs.c), r(rhs.r) { }
     Circle(const Point& c, const double& r): c(c), r(r) { }

     Point point(double ang) const
     {
         return Point(c.x + cos(ang)*r, c.y + sin(ang)*r);    //圆心角所对应的点
     }
     double length(void) const
     {
         return PI * 2.0 * r;
     }
     double area(void) const
     {
         return PI * r * r;
     }
     void in(void)const
     {
         scanf("%lf%lf%lf",&c.x,&c.y,&r);
     }
 };
 double CCdistance(Circle a, Circle b)//圆心距
 {
     return sqrt((a.c.x-b.c.x)*(a.c.x-b.c.x)+(a.c.y-b.c.y)*(a.c.y-b.c.y));
 }
 int CCguanxi(Circle a,Circle b)//两圆关系
 {
     double dis=CCdistance(a,b);
     if(dis>=a.r+b.r)
         return -;  //外离||外切
     else if(dis<fabs(a.r-b.r))
         return ;   //内含
     else if(fabs(dis-fabs(a.r-b.r))<eps)
         return ; //内切
     else
         return ;//相交
 }
 double CCyuanxinjiao(Circle a,Circle b) //a,b相交 a的圆心角
 {
     double dis=CCdistance(a,b);
     double temp=(dis*dis+a.r*a.r-b.r*b.r)/(*dis*a.r);
     return 2.0*acos(temp);
 }
 double Arclength(Circle a,double jiaodu) //圆心角所对应的弧长
 {
     return jiaodu*a.r;
 }
 Circle c[maxn];
 Point p[maxn];
 int main()
 {
     int t;
     double r,m;
     cin>>t;
     while(t--)
     {
         cin>>m>>r;
         Circle yuan(Point(,),r);
         double ans=yuan.length();
         for(int i=;i<m;i++)
             c[i].in();
         for(int i=;i<m;i++)
         {
             //cout<<i<<" "<<c[i].zhouchang()<<endl;
             if(CCguanxi(yuan,c[i])==)
                 ans+=c[i].length();
             else if(CCguanxi(yuan,c[i])==)
             {
                 double a=CCyuanxinjiao(yuan,c[i]);
                 ans-=Arclength(yuan,a);
                 double b=CCyuanxinjiao(c[i],yuan);
                 ans+=Arclength(c[i],b);
             }
         }
         printf("%.20f\n",ans);
     }
 }