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 ≤ 10 ¹²)
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：求n个m形能将空间最多化成几部分。
思路：每个m形都尽量去穿过前n-1个m，推出公式得：8*n^2-7*n+1,利用C++大数模板处理，加上输入外挂，Java卡时间。

#include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; typedef long long ll; /* * 完全大数模板 * 输出cin>>a * 输出a.print(); * 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。 * by kuangbin GG. */ #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;memset(a,0,sizeof(a));} //构造函数 BigNum(const ll); //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 ll b) //将一个int类型的变量转化为大数 { ll 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
       
        >(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<
        
         =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
         
          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
          
           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
           
            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/(con