Sawtooth
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 979 Accepted Submission(s): 375
Problem Description
Think about a plane:
● One straight line can divide a plane into two regions.
● Two lines can divide a plane into at most four regions.
● Three lines can divide a plane into at most seven regions.
● And so on...
Now we have some figure constructed with two parallel rays in the same direction, joined by two straight segments. It looks like a character “M”. You are given N such “M”s. What is the maximum number of regions that these “M”s can divide a plane ?
Input
The first line of the input is T (1 ≤ T ≤ 100000), which stands for the number of test cases you need to solve.
Each case contains one single non-negative integer, indicating number of “M”s. (0 ≤ N ≤ 1012)
Output
For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then an integer that is the maximum number of regions N the “M” figures can divide.
Sample Input
2 1 2
Sample Output
Case #1: 2 Case #2: 19
题意:n个M型(有两条要平行),最多能将整个空间分成多少份。
思路:首先考虑简单的情况:即n条直线最多能将空间分成多少份,用L(n)来表示
L(0) = 1,L(1) = 2,L(2) = 4,L(3) = 7........
很容易想到,求L(n)时,这一条直线最多与n-1条直线有n-1个交点,增加了n个区域(对n个已有区域进行了划分)因此递推式就为:
L(n) = L(n-1)+n
可以得到L(n) = (n+1)*n/2+1
再过来看M型:
如果把一个M型看成4条直线(两条边平行),那么我们可以直接通过刚才那个式子得到答案!
但是,它是射线!如果尝试着把线段无限延长,就会发现其实你可以把这个M型看成是4条直线划分空间但是有些空间是浪费的(这些浪费的空间都是在那三个头),而且可以算出损失的空间数为9,但是容易想到你想要划分的空间最大,就是要让每个M型的三个头都露在外面!(贪心)这样的话,因此每个M型的三个头都露在外面,因此可以得到公式
M(n) = L(4*n)-9*n = 8n^2-7n+1
另外,如果两边边不平行,结果也是一样的 !
代码细节优化就没啥意思了。。。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; #define MAXN 9999 #define MAXSIZE 1010 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; public: BigNum() { len=1; //构造函数 memset(a,0,sizeof(a)); } BigNum(const long long); //将一个int类型的变量转化成大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符,大数之间进行赋值运算 friend istream& operator>>(istream&,BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符 BigNum operator+(const BigNum &)const; //重载加法运算符,两个大数之间的相加运算 BigNum operator-(const BigNum &)const; //重载减法运算符,两个大数之间的相减运算 BigNum operator*(const BigNum &)const; //重载乘法运算符,两个大数之间的相乘运算 BigNum operator/(const int &)const; //重载除法运算符,大数对一个整数进行相除运算 BigNum operator^(const int &)const; //大数的n次方运算 int operator%(const int &)const; //大数对一个int类型的变量进行取模运算 bool operator>(const BigNum &T)const; //大数和另一个大数的大小比较 bool operator>(const int &t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const long long b) //将一个int类型的变量转化为大数 { long long c,d=b; len=0; memset(a,0,sizeof(a)); while(d>MAXN) { c=d-(d/(MAXN+1))*(MAXN+1); d=d/(MAXN+1); a[len++]=c; } a[len++]=d; } BigNum::BigNum(const char *s) //将一个字符串类型的变量转化为大数 { int t,k,index,L,i; memset(a,0,sizeof(a)); L=strlen(s); len=L/DLEN; if(L%DLEN)len++; index=0; for(i=L-1; i>=0; i-=DLEN) { t=0; k=i-DLEN+1; if(k<0)k=0; for(int j=k; j<=i; j++) t=t*10+s[j]-'0'; a[index++]=t; } } BigNum::BigNum(const BigNum &T):len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i=0; i<len; i++) a[i]=T.a[i]; } BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符,大数之间赋值运算 { int i; len=n.len; memset(a,0,sizeof(a)); for(i=0; i<len; i++) a[i]=n.a[i]; return *this; } istream& operator>>(istream &in,BigNum &b) { char ch[MAXSIZE*4]; int i=-1; in>>ch; int L=strlen(ch); int count=0,sum=0; for(i=L-1; i>=0;) { sum=0; int t=1; for(int j=0; j<4&&i>=0; j++,i--,t*=10) { sum+=(ch[i]-'0')*t; } b.a[count]=sum; count++; } b.len=count++; return in; } ostream& operator<<(ostream& out,BigNum& b) //重载输出运算符 { int i; cout<<b.a[b.len-1]; for(i=b.len-2; i>=0; i--) { printf("%04d",b.a[i]); } return out; } BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算 { BigNum t(*this); int i,big; big=T.len>len?T.len:len; for(i=0; i<big; i++) { t.a[i]+=T.a[i]; if(t.a[i]>MAXN) { t.a[i+1]++; t.a[i]-=MAXN+1; } } if(t.a[big]!=0) t.len=big+1; else t.len=big; return t; } BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算 { int i,j,big; bool flag; BigNum t1,t2; if(*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for(i=0; i<big; i++) { if(t1.a[i]<t2.a[i]) { j=i+1; while(t1.a[j]==0) j++; t1.a[j--]--; while(j>i) t1.a[j--]+=MAXN; t1.a[i]+=MAXN+1-t2.a[i]; } else t1.a[i]-=t2.a[i]; } t1.len=big; while(t1.a[len-1]==0 && t1.len>1) { t1.len--; big--; } if(flag) t1.a[big-1]=0-t1.a[big-1]; return t1; } BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘 { BigNum ret; int i,j,up; int temp,temp1; for(i=0; i<len; i++) { up=0; for(j=0; j<T.len; j++) { temp=a[i]*T.a[j]+ret.a[i+j]+up; if(temp>MAXN) { temp1=temp-temp/(MAXN+1)*(MAXN+1); up=temp/(MAXN+1); ret.a[i+j]=temp1; } else { up=0; ret.a[i+j]=temp; } } if(up!=0) ret.a[i+j]=up; } ret.len=i+j; while(ret.a[ret.len-1]==0 && ret.len>1)ret.len--; return ret; } BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算 { BigNum ret; int i,down=0; for(i=len-1; i>=0; i--) { ret.a[i]=(a[i]+down*(MAXN+1))/b; down=a[i]+down*(MAXN+1)-ret.a[i]*b; } ret.len=len; while(ret.a[ret.len-1]==0 && ret.len>1) ret.len--; return ret; } int BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模 { int i,d=0; for(i=len-1; i>=0; i--) d=((d*(MAXN+1))%b+a[i])%b; return d; } bool BigNum::operator>(const BigNum &T)const //大数和另一个大数的大小比较 { int ln; if(len>T.len)return true; else if(len==T.len) { ln=len-1; while(a[ln]==T.a[ln]&&ln>=0) ln--; if(ln>=0 && a[ln]>T.a[ln]) return true; else return false; } else return false; } bool BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; printf("%d",a[len-1]); for(i=len-2; i>=0; i--) printf("%04d",a[i]); printf("\n"); } int main() { int ncase,T=1; cin >> ncase; while(ncase--) { long long t; scanf("%I64d",&t); if(t <= 1e9) { printf("Case #%d: %I64d\n",T++,8*t*t-7*t+1); } else { BigNum tmp = BigNum(t); printf("Case #%d: ",T++); tmp = (tmp*tmp*8-tmp*7+1); tmp.print(); } } return 0; }