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+50,inf=1e18;
const ll mod=1e9+7;
vector<int> per[10][10];
int vis[11],a[10],b[10];
void init()
{
    for(int i=1;i<=9;i++)
    {
        for(int j=1;j<=i;j++)
            a[j]=j;
        do
        {
            fillchar(vis,0);
            int now=0,temp=0;
            for(int j=1;j<=i;j++)
                now=now*10+a[j];
            for(int j=1;j<=i;j++)
            {
                if(vis[j]==0)
                {
                    vis[j]=1;
                    for(int k=a[j];k!=j;k=a[k])
                        vis[k]=1,temp++;
                }
            }
            per[i][temp].pb(now);
        }while(next_permutation(a+1,a+i+1));
    }
}
char s[12];
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==10)
        {
            printf("%d %d\n",minn,maxx);
            continue;
        }
        reverse(s,s+len);
        if(len<=k)k=len-1;
        for(int i=0;i<=k;i++)
        {
            for(int j=0;j<per[len][i].size();j++)
            {
                int temp=per[len][i][j],cnt=0,sum=0;
                while(temp)
                {
                   sum=sum*10+int(s[temp%10-1]-'0');
                   temp/=10;
                }
                //cout<<per[len][i][j]<<" "<<sum<<endl;
                if((int)pow(10,len-1)>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

  1 #include <bits/stdc++.h>
  2 #define LL long long
  3 #define PI 3.1415926535897932384626
  4 #define maxn 1000
  5 #define EXIT exit(0);
  6 #define DEBUG puts("Here is a BUG");
  7 #define CLEAR(name, init) memset(name, init, sizeof(name))
  8 const double eps = 1e-9;
  9 const int MAXN = (int)1e9 + 5;
 10 using namespace std;
 11 #define Vector Point
 12 int dcmp(double x)
 13 {
 14     return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1);
 15 }
 16 struct Point
 17 {
 18     double x, y;
 19 
 20     Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数
 21     Point(double x = 0.0, double y = 0.0): x(x), y(y) { }   //构造函数
 22     friend istream& operator >> (istream& in, Point& P)
 23     {
 24         return in >> P.x >> P.y;
 25     }
 26     friend ostream& operator << (ostream& out, const Point& P)
 27     {
 28         return out << P.x << ' ' << P.y;
 29     }
 30     friend Vector operator + (const Vector& A, const Vector& B)
 31     {
 32         return Vector(A.x+B.x, A.y+B.y);
 33     }
 34     friend Vector operator - (const Point& A, const Point& B)
 35     {
 36         return Vector(A.x-B.x, A.y-B.y);
 37     }
 38     friend Vector operator * (const Vector& A, const double& p)
 39     {
 40         return Vector(A.x*p, A.y*p);
 41     }
 42     friend Vector operator / (const Vector& A, const double& p)
 43     {
 44         return Vector(A.x/p, A.y/p);
 45     }
 46     friend bool operator == (const Point& A, const Point& B)
 47     {
 48         return dcmp(A.x-B.x) == 0 && dcmp(A.y-B.y) == 0;
 49     }
 50     friend bool operator < (const Point& A, const Point& B)
 51     {
 52         return A.x < B.x || (A.x == B.x && A.y < B.y);
 53     }
 54     void in(void)
 55     {
 56         scanf("%lf%lf", &x, &y);
 57     }
 58     void out(void)
 59     {
 60         printf("%lf %lf", x, y);
 61     }
 62 };
 63 template <class T> T sqr(T x)
 64 {
 65     return x * x;
 66 }
 67 double Dot(const Vector& A, const Vector& B)
 68 {
 69     return A.x*B.x + A.y*B.y;    //点积
 70 }
 71 double Length(const Vector& A)
 72 {
 73     return sqrt(Dot(A, A));
 74 }
 75 double Angle(const Vector& A, const Vector& B)
 76 {
 77     return acos(Dot(A, B)/Length(A)/Length(B));    //向量夹角
 78 }
 79 double Cross(const Vector& A, const Vector& B)
 80 {
 81     return A.x*B.y - A.y*B.x;    //叉积
 82 }
 83 double Area(const Point& A, const Point& B, const Point& C)
 84 {
 85     return fabs(Cross(B-A, C-A));
 86 }
 87 Vector normal(Vector x)
 88 {
 89     return Point(-x.y, x.x) / Length(x);
 90 }
 91 double angle(Vector x)
 92 {
 93     return atan2(x.y, x.x);
 94 }
 95 Vector vecunit(Vector x)
 96 {
 97     return x / Length(x);   //单位向量
 98 }
 99 struct Circle
100 {
101     Point c;    //圆心
102     double r;   //半径
103     Circle() { }
104     Circle(const Circle& rhs): c(rhs.c), r(rhs.r) { }
105     Circle(const Point& c, const double& r): c(c), r(r) { }
106 
107     Point point(double ang) const
108     {
109         return Point(c.x + cos(ang)*r, c.y + sin(ang)*r);    //圆心角所对应的点
110     }
111     double length(void) const
112     {
113         return PI * 2.0 * r;
114     }
115     double area(void) const
116     {
117         return PI * r * r;
118     }
119     void in(void)const
120     {
121         scanf("%lf%lf%lf",&c.x,&c.y,&r);
122     }
123 };
124 double CCdistance(Circle a, Circle b)//圆心距
125 {
126     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));
127 }
128 int CCguanxi(Circle a,Circle b)//两圆关系
129 {
130     double dis=CCdistance(a,b);
131     if(dis>=a.r+b.r)
132         return -1;  //外离||外切
133     else if(dis<fabs(a.r-b.r))
134         return 0;   //内含
135     else if(fabs(dis-fabs(a.r-b.r))<eps)
136         return 2; //内切
137     else
138         return 1;//相交
139 }
140 double CCyuanxinjiao(Circle a,Circle b) //a,b相交 a的圆心角
141 {
142     double dis=CCdistance(a,b);
143     double temp=(dis*dis+a.r*a.r-b.r*b.r)/(2*dis*a.r);
144     return 2.0*acos(temp);
145 }
146 double Arclength(Circle a,double jiaodu) //圆心角所对应的弧长
147 {
148     return jiaodu*a.r;
149 }
150 Circle c[maxn];
151 Point p[maxn];
152 int main()
153 {
154     int t;
155     double r,m;
156     cin>>t;
157     while(t--)
158     {
159         cin>>m>>r;
160         Circle yuan(Point(0,0),r);
161         double ans=yuan.length();
162         for(int i=0;i<m;i++)
163             c[i].in();
164         for(int i=0;i<m;i++)
165         {
166             //cout<<i<<" "<<c[i].zhouchang()<<endl;
167             if(CCguanxi(yuan,c[i])==2)
168                 ans+=c[i].length();
169             else if(CCguanxi(yuan,c[i])==1)
170             {
171                 double a=CCyuanxinjiao(yuan,c[i]);
172                 ans-=Arclength(yuan,a);
173                 double b=CCyuanxinjiao(c[i],yuan);
174                 ans+=Arclength(c[i],b);
175             }
176         }
177         printf("%.20f\n",ans);
178     }
179 }
View Code

 

posted @ 2018-08-07 16:45  灬从此以后灬  阅读(210)  评论(0)  编辑  收藏  举报