13级个人结业赛-数学场

数学场题解。(现学现卖，有缺漏之处，还望包涵。)

从题目的分布上来看，A,B,C为数论内容,DFI为计算几何,E为矩阵,G题概率。希望你们在本次比赛中有所收获。

数学A

对于传统的ax+by=gcd(a,b)使用扩展欧几里得解即可,但是这边题面是ax+by+c=0

必然存在着差异。当我们求出ax+by=gcd(x,y)题解(x0,y0)的时候,ax+by+c=0的解就是

( x0*(-c/gcd(a,b)), y0*(-c/gcd(a,b))); 假如c%gcd(a,b)!=0那么算出来的解就不是整数解了，也就是无解的情况。

这边给出一个博客供大家学习:http://blog.csdn.net/daniel_ustc/article/details/8273201

#include <iostream>  
#include <stdio.h>  
#include <stdlib.h>  
#include <math.h>  
#include <string.h>  
#include <algorithm>  
using namespace std;  
typedef long long ll;  
const ll maxn=5000000000000000000;  
int is(ll a,ll b)  
{  
    if(fabs(a)<=maxn&&fabs(b)<=maxn) return 1;  
    return 0;  
}  
ll gcd_ex(ll a,ll b,ll *x,ll *y)  
{  
    if(b==0)  
    {  
        *y=0;  
        *x=1;  
        return a;  
    }  
    ll r=gcd_ex(b,a%b,x,y);  
    ll t=*x;  
    *x=*y;  
    *y=t-a/b*(*y);  
    return r;  
}  
    
int main()  
{  
    ll a,b,c,x,y,gcdans;  
    while(scanf("%lld%lld%lld",&a,&b,&c)!=EOF)  
    {  
        gcdans=gcd_ex(a,b,&x,&y);  
        if(c%gcdans==0)  
        {  
            x*=(-c)/gcdans;  
            y*=(-c)/gcdans;  
        }  
        else
        {  
            printf("%lld %lld
",maxn+1,maxn+1);  
            continue;  
        }  
        printf("%lld %lld
",x,y);  
    }  
    return 0;  
}  

　　　　　　　　　　　　　　　　　　　　　　　　　　数学B

求n=a*b*c因子个数。首先对n进行质因数分解。比如n=p1^x1*p2^x2….pk^xk  ,p1,p2,p3…pk为质数。

那么他的因子就会有(x1+1)*(x2+1)*…*(xk+1)种组合；记得记忆化。不然会超时。

#include <iostream> 
#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 
#include <string.h> 
#include <algorithm> 
using namespace std; 
typedef long long ll; 
const int maxn=1000005; 
ll a,b,c,num[maxn]= {0},prime[maxn]= {0},status[maxn]= {0}; 
const ll mod=1073741824; 
int workprime() 
{ 
    int ans=0; 
    for(int i=2; i<=maxn; i++) 
    { 
        if(!status[i]) 
        { 
            prime[ans++]=i; 
            status[i]=1; 
            for(int j=i; j<=maxn; j+=i) 
                status[j]=1; 
        } 
    } 
    return ans; 
} 
void count(ll r) 
{ 
    int ans=r; 
    if(!num[r]) 
    { 
        int b=0; 
        num[ans]=1; 
        while(r>=prime[b]) 
        { 
            int cou=0; 
            while(r%prime[b]==0) 
            { 
                cou+=1; 
                r/=prime[b]; 
            } 
            b++; 
            num[ans]=num[ans]*(cou+1); 
            if(r==1) break; 
        } 
    } 
} 
int main() 
{ 
    ll sum=0,len; 
  
    memset(prime,0,sizeof(prime)); 
    len=workprime(); 
    memset(num,0,sizeof(num)); 
    while(scanf("%lld%lld%lld",&a,&b,&c)!=EOF) 
    { 
        sum=0; 
        for(ll i=1; i<=a; i++) 
        { 
            for(ll j=1; j<=b; j++) 
            { 
                for(ll k=1; k<=c; k++) 
                { 
                    count(i*j*k); 
                    sum=(sum%mod+num[i*j*k])%mod; 
                } 
            } 
        } 
        printf("%lld
",sum); 
    } 
    return 0; 
} 

数学C

先求出x,y的最简比例。然后求出可以扩大的最大比例k,对应的答案就是k*x,k*y。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a,b,x,y;
int main()
{
    while(scanf("%d%d%d%d",&a,&b,&x,&y)!=EOF)
    {
        int ans=__gcd(x,y);
        x/=ans;
        y/=ans;
        int k=min(a/x,b/y);///找出允许的最大比例
        printf("%d %d
",x*k,y*k);///从最终的比例上扩大倍数
    }
    return 0;
}

数学D

求出两个矩形的交点。在交点中可能出现4交点，也可能出现8交点。这个就取决于a的大小，当然我们需要求出一个临界角度，旋转之后矩形的对角线刚好和未旋转的矩形对角线重合。求出角度后再求出点，再加个凸包的面积计算就可以了。(听说还可以用”半平面交”然而我不会)

#include <iostream> 
#include <iomanip> 
#include <stdio.h> 
#include <set> 
#include <vector> 
#include <map> 
#include <cmath> 
#include <algorithm> 
#include <memory.h> 
#include <string> 
#include <sstream> 
  
using namespace std; 
  
const long double pi = 3.1415926535897932384626433832795; 
  
long double x[42], y[42]; 
int n; 
  
void add(int qx, int qy, long double ang) 
{ 
    x[n] = cos(ang)*qx-sin(ang)*qy; 
    y[n] = sin(ang)*qx+cos(ang)*qy; 
    n++; 
} 
  
void cut(long double xa, long double ya, long double xb, long double yb) 
{ 
    long double a = yb-ya, b = xa-xb; 
    long double c = -a*xa-b*ya; 
    long double eps = (1e-10)*1.00042; 
    int nn = 0; 
    long double xx[42], yy[42]; 
    x[n] = x[0]; 
    y[n] = y[0]; 
    for (int i=0; i<n; i++) 
    { 
        long double z1 = a*x[i]+b*y[i]+c; 
        if (z1 < eps) 
        { 
            xx[nn] = x[i]; 
            yy[nn] = y[i]; 
            nn++; 
        } 
        long double z2 = a*x[i+1]+b*y[i+1]+c; 
        if (z1 < eps && z2 > eps || z1 > eps && z2 < eps) 
        { 
            long double aa = y[i+1]-y[i], bb = x[i]-x[i+1]; 
            long double cc = -aa*x[i]-bb*y[i]; 
            long double d = a*bb-b*aa; 
            xx[nn] = (b*cc-c*bb)/d; 
            yy[nn] = (c*aa-a*cc)/d; 
            nn++; 
        } 
    } 
    n = nn; 
    for (int i=0; i<n; i++) x[i] = xx[i], y[i] = yy[i]; 
} 
  
int main() 
{ 
    //freopen("in.txt","r",stdin); 
    //freopen("out.txt","w",stdout); 
    int w, h, iang; 
    while(scanf("%d %d %d", &w, &h, &iang)!=EOF) 
    { 
        long double ang = iang/180.0*pi; 
        n = 0; 
        add(w, h, ang); 
        add(-w, h, ang); 
        add(-w, -h, ang); 
        add(w, -h, ang); 
        cut(w, h, -w, h); 
        cut(-w, h, -w, -h); 
        cut(-w, -h, w, -h); 
        cut(w, -h, w, h); 
        x[n] = x[0]; 
        y[n] = y[0]; 
        long double area = 0; 
        for (int i=0; i<n; i++) area += (x[i]-x[i+1])*(y[i]+y[i+1]); 
        printf("%f
", (double)(0.125*area)); 
    } 
    return 0; 
} 

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
const double eps=1e-10;
const double pi=acos(-1.0);
double  w,h,a;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}///11?¨¬o¡¥¨ºy
};
Point p[10];
typedef Point VC;
///向量+向量=向量,点+向量=点
VC operator +(VC A,VC B)
{
    return VC(A.x+B.x,A.y+B.y);
}
///点-点=向量
VC operator -(VC A,VC B)
{
    return VC(A.x-B.x,A.y-B.y);
}
///向量*数=向量
VC operator *(VC A,double p)
{
    return VC(A.x*p,A.y*p);
}
///向量/数=向量
VC operator /(VC A,double p)
{
    return VC(A.x/p,A.y/p);
}
double Cross(VC A,VC B)
{
    return A.x*B.y-A.y*B.x;
}
double Area2(Point A,Point B,Point C)
{
    return Cross(B-A,C-A);
}
VC Rotate(VC A,double rad )
{
    return VC(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
Point GetLineIntersection(Point P,VC v,Point Q,VC w)
{
    VC u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
double ConvexPolygonArea(Point *p,int n)
{
    double area=0;
    for(int i=1; i<n-1; i++)
        area+=Cross(p[i]-p[0],p[i+1]-p[0]);
    return area/2;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out2.txt","w",stdout);
    while(scanf("%lf%lf%lf",&w,&h,&a)!=EOF)
    {

        if(h<w) h=w+h,w=h-w,h=h-w;
        int len=0;
        if(a==0||a==180)
        {
            printf("%f
",w*h);
            continue;
        }
        Point A,AA,B,BB,C,CC,D,DD;
        if(a>90) a=180-a;
        A=Point(h/2,w/2);
        AA=Rotate(A,a*pi/180);
        B=Point(-h/2,w/2);
        BB=Rotate(B,a*pi/180);
        C=Point(-h/2,-w/2);
        CC=Rotate(C,a*pi/180);
        D=Point(h/2,-w/2);
        DD=Rotate(D,a*pi/180);
        if(a*1.0-2*(atan(1.0*w/h)*180/pi)>0)
        {

            Point p1=GetLineIntersection(A,B-A,DD,CC-DD);
            Point p2=GetLineIntersection(A,B-A,AA,BB-AA);
            Point p3=GetLineIntersection(D,C-D,DD,CC-DD);
             printf("%f
",Area2(p1,p2,p3));
        }
        else if(fabs(a*1.0-2*(atan(1.0*w/h)*180/pi))<eps)
        {
            printf("%f
",w*h);
            continue;
        }
        else
        {
            p[0]=GetLineIntersection(A,B-A,AA,DD-AA);
            p[1]=GetLineIntersection(A,B-A,AA,BB-AA);
            p[2]=GetLineIntersection(B,C-B,BB,AA-BB);
            p[3]=GetLineIntersection(B,C-B,BB,CC-BB);
            p[4]=GetLineIntersection(C,D-C,CC,BB-CC);
            p[5]=GetLineIntersection(C,D-C,CC,DD-CC);
            p[6]=GetLineIntersection(D,A-D,DD,CC-DD);
            p[7]=GetLineIntersection(D,A-D,DD,AA-DD);
            printf("%f
",ConvexPolygonArea(p,8));
        }
    }
}

数学E

由于n的过大。线性的求法难以忍受。矩阵快速幂+取模(忘记了我就呵呵哒),就可以解决。

http://www.cnblogs.com/vongang/archive/2012/04/01/2429015.html

#include <iostream> 
#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <algorithm> 
using namespace std; 
int aa,bb,n,m,mod[]={1,10,100,1000,10000,100000}; 
struct mxt 
{ 
    int v[2][2]; 
}ans,temp; 
mxt mul(mxt a,mxt b) 
{ 
    mxt en; 
    memset(en.v,0,sizeof(en.v)); 
    for(int i=0;i<2;i++) 
        for(int j=0;j<2;j++) 
        for(int k=0;k<2;k++) 
        en.v[i][j]=(en.v[i][j]+a.v[i][k]*b.v[k][j])%mod[m]; 
    return en; 
} 
mxt pow_mod(mxt a,int k) 
{ 
    mxt r; 
    memset(r.v,0,sizeof(r.v)); 
    r.v[0][0]=r.v[1][1]=1;
    while(k) 
    { 
        if(k&1) r=mul(r,a); 
        a=mul(a,a); 
        k>>=1; 
    } 
    return r; 
} 
  
int main() 
{ 
    int t; 
    scanf("%d",&t); 
    while(t--) 
    { 
        scanf("%d%d%d%d",&aa,&bb,&n,&m); 
        if(n==0) printf("%d
",aa%mod[m]); 
        else if(n==1) printf("%d
",bb%mod[m]); 
        else
        { 
            memset(temp.v,0,sizeof(temp.v)); 
            temp.v[0][0]=aa; 
            temp.v[0][1]=bb; 
            temp.v[1][0]=aa+bb; 
            temp.v[1][1]=bb; 
            ans.v[0][0]=ans.v[0][1]=ans.v[1][0]=1;ans.v[1][1]=0; 
            ans=pow_mod(ans,n-1); 
            printf("%d
",(ans.v[0][0]*bb+ans.v[0][1]*aa)%mod[m]); 
        } 
    } 
    return 0; 
} 

数学F

枚举出四边形的对角线，然后分别在对角线两边(使用叉积判断点在对角线周围的位置)找距离对角线最远的点。要注意对于某一条对角线,如果其他点都在他的一侧的话，就算这个面积再大，也是不能取的

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=305;
const double eps=1e-10;
const double pi=acos(-1);
int  n;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}///构造函数
};
Point p[maxn];
typedef Point VC;
///向量+向量=向量,点+向量=点
VC operator +(VC A,VC B){return VC(A.x+B.x,A.y+B.y);}
///点-点=向量
VC operator -(VC A,VC B) {return VC(A.x-B.x,A.y-B.y);}
///向量*数=向量
VC operator *(VC A,double p) {return VC(A.x*p,A.y*p);}
///向量/数=向量
VC operator /(VC A,double p) {return VC(A.x/p,A.y/p);}
double Cross(VC A,VC B){return A.x*B.y-A.y*B.x;}
double Dot(VC A,VC B) {return A.x+B.x+A.y*B.y;}///点积=|A|*|B|*cos<A,B>,A，B的逆时针转旋角,满足交换率
double length(VC A) {return sqrt(Dot(A,A));}///长度
VC Rotate(VC A,double rad)///rad弧度
{
    return VC(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
Point GetLineIntersection(Point P,VC v,Point Q,VC w)
{
    VC u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
double DistanceToLine(Point P,Point A,Point B)///p到直线a,b
{
    VC v1=B-A,v2=P-A;
    return fabs(Cross(v1,v2))/length(v1);///如不取绝对值，得到的是又向距离。
}
int dcmp(double x)///涉及精度问题
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
bool OnSegment(Point p,Point a1,Point a2)
{
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
VC vv[maxn*maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    int len=0;
    while(scanf("%d",&n)!=EOF)
    {
        //printf("%d
",n);
        len=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        double  ansmax=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                //vv[len]=VC(p[j]-p[i]);///向量起点为p[i]
                double ansmax1,ansmax2,anslength=length(p[j]-p[i]);
                //printf("length=%f
",anslength);
                ansmax1=ansmax2=0;
                for(int k=0;k<n;k++)
                {
                    if(k==i||k==j) continue;
                    else
                    {
                        double ans=Cross(p[j]-p[i],p[k]-p[i]);
                        if(ans>0) ansmax1=max(ansmax1,DistanceToLine(p[k],p[i],p[j]));
                        else if(ans<0) ansmax2=max(ansmax2,DistanceToLine(p[k],p[i],p[j]));
                    }
                }
                if(ansmax1!=0&&ansmax2!=0)
                ansmax=max(ansmax,anslength*(ansmax1+ansmax2)/2);
            }
        }
        printf("%f
",ansmax);
    }
}

数学G

裸尼姆博弈，答案在必胜态和必败态里面。

http://www.cnblogs.com/kuangbin/archive/2011/08/28/2156426.html

#include <iostream> 
#include <stdio.h> 
#include <stdlib.h> 
#include <algorithm> 
using namespace std; 
int main() 
{ 
    int t; 
    scanf("%d",&t); 
    while(t--) 
    { 
        int n; 
        scanf("%d",&n); 
        int flag=0;
        int s=0; 
        for(int i=0;i<n;i++) 
        { 
            int x; 
            scanf("%d",&x); 
            if(i==0) s=x; 
            else s^=x; 
            if(x>1) flag=1; 
        } 
        if(flag==0) 
        { 
            if(n%2==0) {printf("A
");continue;} 
            printf("B
"); 
        } 
        else
        { 
            if(s==0) {printf("B
");continue;} 
            printf("A
"); 
        } 
    } 
    return 0; 
} 

数学H

最大值的分布是1,2,….n。分别求出对应的可能数量。比如最大值是k(k>1)，那么以k为最大值的数量有k^n-(k-1)^n.然后我们就可以求出概率了。E=概率*k.对E进行化简就会得出E=m-(1/m)^n-(2/m)^n-…-((m-1)/m)^n

#include<stdio.h> 
#include<math.h> 
int main() 
{ 
    int n,m; 
    double sum; 
    while(scanf("%d%d",&m,&n)!=EOF) 
    { 
        sum=m; 
        for(int i=1; i<m; i++) 
        { 
            sum-=pow((double)1.0-(double)i/m,(double)n); 
        } 
        printf("%.12lf
",sum); 
    } 
    return 0; 
}