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
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <string.h> 6 #include <algorithm> 7 using namespace std; 8 typedef long long ll; 9 const ll maxn=5000000000000000000; 10 int is(ll a,ll b) 11 { 12 if(fabs(a)<=maxn&&fabs(b)<=maxn) return 1; 13 return 0; 14 } 15 ll gcd_ex(ll a,ll b,ll *x,ll *y) 16 { 17 if(b==0) 18 { 19 *y=0; 20 *x=1; 21 return a; 22 } 23 ll r=gcd_ex(b,a%b,x,y); 24 ll t=*x; 25 *x=*y; 26 *y=t-a/b*(*y); 27 return r; 28 } 29 30 int main() 31 { 32 ll a,b,c,x,y,gcdans; 33 while(scanf("%lld%lld%lld",&a,&b,&c)!=EOF) 34 { 35 gcdans=gcd_ex(a,b,&x,&y); 36 if(c%gcdans==0) 37 { 38 x*=(-c)/gcdans; 39 y*=(-c)/gcdans; 40 } 41 else 42 { 43 printf("%lld %lld\n",maxn+1,maxn+1); 44 continue; 45 } 46 printf("%lld %lld\n",x,y); 47 } 48 return 0; 49 }
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。
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <string.h> 6 #include <algorithm> 7 using namespace std; 8 typedef long long ll; 9 const int maxn=1000005; 10 ll a,b,c,num[maxn]= {0},prime[maxn]= {0},status[maxn]= {0}; 11 const ll mod=1073741824; 12 int workprime() 13 { 14 int ans=0; 15 for(int i=2; i<=maxn; i++) 16 { 17 if(!status[i]) 18 { 19 prime[ans++]=i; 20 status[i]=1; 21 for(int j=i; j<=maxn; j+=i) 22 status[j]=1; 23 } 24 } 25 return ans; 26 } 27 void count(ll r) 28 { 29 int ans=r; 30 if(!num[r]) 31 { 32 int b=0; 33 num[ans]=1; 34 while(r>=prime[b]) 35 { 36 int cou=0; 37 while(r%prime[b]==0) 38 { 39 cou+=1; 40 r/=prime[b]; 41 } 42 b++; 43 num[ans]=num[ans]*(cou+1); 44 if(r==1) break; 45 } 46 } 47 } 48 int main() 49 { 50 ll sum=0,len; 51 52 memset(prime,0,sizeof(prime)); 53 len=workprime(); 54 memset(num,0,sizeof(num)); 55 while(scanf("%lld%lld%lld",&a,&b,&c)!=EOF) 56 { 57 sum=0; 58 for(ll i=1; i<=a; i++) 59 { 60 for(ll j=1; j<=b; j++) 61 { 62 for(ll k=1; k<=c; k++) 63 { 64 count(i*j*k); 65 sum=(sum%mod+num[i*j*k])%mod; 66 } 67 } 68 } 69 printf("%lld\n",sum); 70 } 71 return 0; 72 }
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。
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <string.h> 6 #include <algorithm> 7 using namespace std; 8 int a,b,x,y; 9 int main() 10 { 11 while(scanf("%d%d%d%d",&a,&b,&x,&y)!=EOF) 12 { 13 int ans=__gcd(x,y); 14 x/=ans; 15 y/=ans; 16 int k=min(a/x,b/y);///找出允许的最大比例 17 printf("%d %d\n",x*k,y*k);///从最终的比例上扩大倍数 18 } 19 return 0; 20 }
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)
1 #include <iostream> 2 #include <iomanip> 3 #include <stdio.h> 4 #include <set> 5 #include <vector> 6 #include <map> 7 #include <cmath> 8 #include <algorithm> 9 #include <memory.h> 10 #include <string> 11 #include <sstream> 12 13 using namespace std; 14 15 const long double pi = 3.1415926535897932384626433832795; 16 17 long double x[42], y[42]; 18 int n; 19 20 void add(int qx, int qy, long double ang) 21 { 22 x[n] = cos(ang)*qx-sin(ang)*qy; 23 y[n] = sin(ang)*qx+cos(ang)*qy; 24 n++; 25 } 26 27 void cut(long double xa, long double ya, long double xb, long double yb) 28 { 29 long double a = yb-ya, b = xa-xb; 30 long double c = -a*xa-b*ya; 31 long double eps = (1e-10)*1.00042; 32 int nn = 0; 33 long double xx[42], yy[42]; 34 x[n] = x[0]; 35 y[n] = y[0]; 36 for (int i=0; i<n; i++) 37 { 38 long double z1 = a*x[i]+b*y[i]+c; 39 if (z1 < eps) 40 { 41 xx[nn] = x[i]; 42 yy[nn] = y[i]; 43 nn++; 44 } 45 long double z2 = a*x[i+1]+b*y[i+1]+c; 46 if (z1 < eps && z2 > eps || z1 > eps && z2 < eps) 47 { 48 long double aa = y[i+1]-y[i], bb = x[i]-x[i+1]; 49 long double cc = -aa*x[i]-bb*y[i]; 50 long double d = a*bb-b*aa; 51 xx[nn] = (b*cc-c*bb)/d; 52 yy[nn] = (c*aa-a*cc)/d; 53 nn++; 54 } 55 } 56 n = nn; 57 for (int i=0; i<n; i++) x[i] = xx[i], y[i] = yy[i]; 58 } 59 60 int main() 61 { 62 //freopen("in.txt","r",stdin); 63 //freopen("out.txt","w",stdout); 64 int w, h, iang; 65 while(scanf("%d %d %d", &w, &h, &iang)!=EOF) 66 { 67 long double ang = iang/180.0*pi; 68 n = 0; 69 add(w, h, ang); 70 add(-w, h, ang); 71 add(-w, -h, ang); 72 add(w, -h, ang); 73 cut(w, h, -w, h); 74 cut(-w, h, -w, -h); 75 cut(-w, -h, w, -h); 76 cut(w, -h, w, h); 77 x[n] = x[0]; 78 y[n] = y[0]; 79 long double area = 0; 80 for (int i=0; i<n; i++) area += (x[i]-x[i+1])*(y[i]+y[i+1]); 81 printf("%f\n", (double)(0.125*area)); 82 } 83 return 0; 84 }
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <string.h> 6 #include <algorithm> 7 using namespace std; 8 const double eps=1e-10; 9 const double pi=acos(-1.0); 10 double w,h,a; 11 struct Point 12 { 13 double x,y; 14 Point(double x=0,double y=0):x(x),y(y) {}///11?¨¬o¡¥¨ºy 15 }; 16 Point p[10]; 17 typedef Point VC; 18 ///向量+向量=向量,点+向量=点 19 VC operator +(VC A,VC B) 20 { 21 return VC(A.x+B.x,A.y+B.y); 22 } 23 ///点-点=向量 24 VC operator -(VC A,VC B) 25 { 26 return VC(A.x-B.x,A.y-B.y); 27 } 28 ///向量*数=向量 29 VC operator *(VC A,double p) 30 { 31 return VC(A.x*p,A.y*p); 32 } 33 ///向量/数=向量 34 VC operator /(VC A,double p) 35 { 36 return VC(A.x/p,A.y/p); 37 } 38 double Cross(VC A,VC B) 39 { 40 return A.x*B.y-A.y*B.x; 41 } 42 double Area2(Point A,Point B,Point C) 43 { 44 return Cross(B-A,C-A); 45 } 46 VC Rotate(VC A,double rad ) 47 { 48 return VC(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); 49 } 50 Point GetLineIntersection(Point P,VC v,Point Q,VC w) 51 { 52 VC u=P-Q; 53 double t=Cross(w,u)/Cross(v,w); 54 return P+v*t; 55 } 56 double ConvexPolygonArea(Point *p,int n) 57 { 58 double area=0; 59 for(int i=1; i<n-1; i++) 60 area+=Cross(p[i]-p[0],p[i+1]-p[0]); 61 return area/2; 62 } 63 int main() 64 { 65 //freopen("in.txt","r",stdin); 66 //freopen("out2.txt","w",stdout); 67 while(scanf("%lf%lf%lf",&w,&h,&a)!=EOF) 68 { 69 70 if(h<w) h=w+h,w=h-w,h=h-w; 71 int len=0; 72 if(a==0||a==180) 73 { 74 printf("%f\n",w*h); 75 continue; 76 } 77 Point A,AA,B,BB,C,CC,D,DD; 78 if(a>90) a=180-a; 79 A=Point(h/2,w/2); 80 AA=Rotate(A,a*pi/180); 81 B=Point(-h/2,w/2); 82 BB=Rotate(B,a*pi/180); 83 C=Point(-h/2,-w/2); 84 CC=Rotate(C,a*pi/180); 85 D=Point(h/2,-w/2); 86 DD=Rotate(D,a*pi/180); 87 if(a*1.0-2*(atan(1.0*w/h)*180/pi)>0) 88 { 89 90 Point p1=GetLineIntersection(A,B-A,DD,CC-DD); 91 Point p2=GetLineIntersection(A,B-A,AA,BB-AA); 92 Point p3=GetLineIntersection(D,C-D,DD,CC-DD); 93 printf("%f\n",Area2(p1,p2,p3)); 94 } 95 else if(fabs(a*1.0-2*(atan(1.0*w/h)*180/pi))<eps) 96 { 97 printf("%f\n",w*h); 98 continue; 99 } 100 else 101 { 102 p[0]=GetLineIntersection(A,B-A,AA,DD-AA); 103 p[1]=GetLineIntersection(A,B-A,AA,BB-AA); 104 p[2]=GetLineIntersection(B,C-B,BB,AA-BB); 105 p[3]=GetLineIntersection(B,C-B,BB,CC-BB); 106 p[4]=GetLineIntersection(C,D-C,CC,BB-CC); 107 p[5]=GetLineIntersection(C,D-C,CC,DD-CC); 108 p[6]=GetLineIntersection(D,A-D,DD,CC-DD); 109 p[7]=GetLineIntersection(D,A-D,DD,AA-DD); 110 printf("%f\n",ConvexPolygonArea(p,8)); 111 } 112 } 113 }
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
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <string.h> 5 #include <algorithm> 6 using namespace std; 7 int aa,bb,n,m,mod[]={1,10,100,1000,10000,100000}; 8 struct mxt 9 { 10 int v[2][2]; 11 }ans,temp; 12 mxt mul(mxt a,mxt b) 13 { 14 mxt en; 15 memset(en.v,0,sizeof(en.v)); 16 for(int i=0;i<2;i++) 17 for(int j=0;j<2;j++) 18 for(int k=0;k<2;k++) 19 en.v[i][j]=(en.v[i][j]+a.v[i][k]*b.v[k][j])%mod[m]; 20 return en; 21 } 22 mxt pow_mod(mxt a,int k) 23 { 24 mxt r; 25 memset(r.v,0,sizeof(r.v)); 26 r.v[0][0]=r.v[1][1]=1; 27 while(k) 28 { 29 if(k&1) r=mul(r,a); 30 a=mul(a,a); 31 k>>=1; 32 } 33 return r; 34 } 35 36 int main() 37 { 38 int t; 39 scanf("%d",&t); 40 while(t--) 41 { 42 scanf("%d%d%d%d",&aa,&bb,&n,&m); 43 if(n==0) printf("%d\n",aa%mod[m]); 44 else if(n==1) printf("%d\n",bb%mod[m]); 45 else 46 { 47 memset(temp.v,0,sizeof(temp.v)); 48 temp.v[0][0]=aa; 49 temp.v[0][1]=bb; 50 temp.v[1][0]=aa+bb; 51 temp.v[1][1]=bb; 52 ans.v[0][0]=ans.v[0][1]=ans.v[1][0]=1;ans.v[1][1]=0; 53 ans=pow_mod(ans,n-1); 54 printf("%d\n",(ans.v[0][0]*bb+ans.v[0][1]*aa)%mod[m]); 55 } 56 } 57 return 0; 58 }
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
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <string.h> 6 #include <algorithm> 7 using namespace std; 8 const int maxn=305; 9 const double eps=1e-10; 10 const double pi=acos(-1); 11 int n; 12 struct Point 13 { 14 double x,y; 15 Point(double x=0,double y=0):x(x),y(y) {}///构造函数 16 }; 17 Point p[maxn]; 18 typedef Point VC; 19 ///向量+向量=向量,点+向量=点 20 VC operator +(VC A,VC B){return VC(A.x+B.x,A.y+B.y);} 21 ///点-点=向量 22 VC operator -(VC A,VC B) {return VC(A.x-B.x,A.y-B.y);} 23 ///向量*数=向量 24 VC operator *(VC A,double p) {return VC(A.x*p,A.y*p);} 25 ///向量/数=向量 26 VC operator /(VC A,double p) {return VC(A.x/p,A.y/p);} 27 double Cross(VC A,VC B){return A.x*B.y-A.y*B.x;} 28 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 29 double length(VC A) {return sqrt(Dot(A,A));}///长度 30 VC Rotate(VC A,double rad)///rad弧度 31 { 32 return VC(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); 33 } 34 Point GetLineIntersection(Point P,VC v,Point Q,VC w) 35 { 36 VC u=P-Q; 37 double t=Cross(w,u)/Cross(v,w); 38 return P+v*t; 39 } 40 double DistanceToLine(Point P,Point A,Point B)///p到直线a,b 41 { 42 VC v1=B-A,v2=P-A; 43 return fabs(Cross(v1,v2))/length(v1);///如不取绝对值，得到的是又向距离。 44 } 45 int dcmp(double x)///涉及精度问题 46 { 47 if(fabs(x)<eps) return 0; 48 return x<0?-1:1; 49 } 50 bool OnSegment(Point p,Point a1,Point a2) 51 { 52 return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0; 53 } 54 VC vv[maxn*maxn]; 55 int main() 56 { 57 //freopen("in.txt","r",stdin); 58 int len=0; 59 while(scanf("%d",&n)!=EOF) 60 { 61 //printf("%d\n",n); 62 len=0; 63 for(int i=0;i<n;i++) 64 { 65 scanf("%lf%lf",&p[i].x,&p[i].y); 66 } 67 double ansmax=0; 68 for(int i=0;i<n;i++) 69 { 70 for(int j=i+1;j<n;j++) 71 { 72 //vv[len]=VC(p[j]-p[i]);///向量起点为p[i] 73 double ansmax1,ansmax2,anslength=length(p[j]-p[i]); 74 //printf("length=%f\n",anslength); 75 ansmax1=ansmax2=0; 76 for(int k=0;k<n;k++) 77 { 78 if(k==i||k==j) continue; 79 else 80 { 81 double ans=Cross(p[j]-p[i],p[k]-p[i]); 82 if(ans>0) ansmax1=max(ansmax1,DistanceToLine(p[k],p[i],p[j])); 83 else if(ans<0) ansmax2=max(ansmax2,DistanceToLine(p[k],p[i],p[j])); 84 } 85 } 86 if(ansmax1!=0&&ansmax2!=0) 87 ansmax=max(ansmax,anslength*(ansmax1+ansmax2)/2); 88 } 89 } 90 printf("%f\n",ansmax); 91 } 92 }
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
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 using namespace std; 6 int main() 7 { 8 int t; 9 scanf("%d",&t); 10 while(t--) 11 { 12 int n; 13 scanf("%d",&n); 14 int flag=0; 15 int s=0; 16 for(int i=0;i<n;i++) 17 { 18 int x; 19 scanf("%d",&x); 20 if(i==0) s=x; 21 else s^=x; 22 if(x>1) flag=1; 23 } 24 if(flag==0) 25 { 26 if(n%2==0) {printf("A\n");continue;} 27 printf("B\n"); 28 } 29 else 30 { 31 if(s==0) {printf("B\n");continue;} 32 printf("A\n"); 33 } 34 } 35 return 0; 36 }
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
1 #include<stdio.h> 2 #include<math.h> 3 int main() 4 { 5 int n,m; 6 double sum; 7 while(scanf("%d%d",&m,&n)!=EOF) 8 { 9 sum=m; 10 for(int i=1; i<m; i++) 11 { 12 sum-=pow((double)1.0-(double)i/m,(double)n); 13 } 14 printf("%.12lf\n",sum); 15 } 16 return 0; 17 }
