HDU - 5047 Sawtooth

HDU - 5047 Sawtooth

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

題意：求n個m形能將空間最多化成幾部分。
思路：每個m形都盡量去穿過前n-1個m，推出公式得：8*n^2-7*n+1,利用C++大數模板處理，加上輸入外掛，Java卡時間。

#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;

/*
 * 完全大數模板
 * 輸出cin>>a
 * 輸出a.print();
 * 注意這個輸入不能自動去掉前導0的，可以先讀入到char數組，去掉前導0，再用構造函數。
 * by kuangbin GG.
 */
#define MAXN 9999
#define MAXSIZE 1010
#define DLEN 4
class BigNum
{
    private:
        int a[500]; //可以控制大數的位數
        int len;
    public:
        BigNum(){len=1;memset(a,0,sizeof(a));} //構造函數
        BigNum(const ll); //將一個int類型的變量轉化成大數
        BigNum(const char*); //將一個字符串類型的變量轉化為大數
        BigNum(const BigNum &); //拷貝構造函數
        BigNum &operator=(const BigNum &); //重載賦值運算符，大數之間進行賦值運算
        friend istream& operator>>(istream&,BigNum&); //重載輸入運算符
        friend ostream& operator<<(ostream&,BigNum&); //重載輸出運算符
        BigNum operator+(const BigNum &)const; //重載加法運算符，兩個大數之間的相加運算
        BigNum operator-(const BigNum &)const; //重載減法運算符，兩個大數之間的相減運算
        BigNum operator*(const BigNum &)const; //重載乘法運算符，兩個大數之間的相乘運算
        BigNum operator/(const int &)const; //重載除法運算符，大數對一個整數進行相除運算

        BigNum operator^(const int &)const; //大數的n次方運算
        int operator%(const int &)const; //大數對一個int類型的變量進行取模運算
        bool operator>(const BigNum &T)const; //大數和另一個大數的大小比較
        bool operator>(const int &t)const; //大數和一個int類型的變量的大小比較
        void print(); //輸出大數
};

BigNum::BigNum(const ll b) //將一個int類型的變量轉化為大數
{
    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) //將一個字符串類型的變量轉化為大數
{
    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) //拷貝構造函數
{
    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) //重載輸出運算符
{
    int i;
    cout<=0;i--)
    {
        printf("%04d",b.a[i]);
    }
    return out;
}
BigNum BigNum::operator+(const BigNum &T)const //兩個大數之間的相加運算
{
    BigNum t(*this);
    int i,big;
    big=T.len>len?T.len:len;
    for(i=0;iMAXN)
        {
            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 //兩個大數之間的相減運算
{
    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;ii)
                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 //兩個大數之間的相乘
{
    BigNum ret;
    int i,j,up;
    int temp,temp1;
    for(i=0;iMAXN)
            {
                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 //大數對一個整數進行相除運算
{
    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 //大數對一個 int類型的變量進行取模
{
    int i,d=0;
    for(i=len-1;i>=0;i--)
        d=((d*(MAXN+1))%b+a[i])%b;
    return d;
}
BigNum BigNum::operator^(const int &n)const //大數的n次方運算
{
    BigNum t,ret(1);
    int i;
    if(n<0)exit(-1);
    if(n==0)return 1;
    if(n==1)return *this;
    int m=n;
    while(m>1)
    {
        t=*this;
        for(i=1;(i<<1)<=m;i<<=1)
            t=t*t;
        m-=i;
        ret=ret*t;
        if(m==1)ret=ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum &T)const //大數和另一個大數的大小比較
{
    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 //大數和一個int類型的變量的大小比較
{
    BigNum b(t);
    return *this>b;
}
void BigNum::print() //輸出大數
{
    int i;
    printf("%d",a[len-1]);
    for(i=len-2;i>=0;i--)
        printf("%04d",a[i]);
    printf("\n");
}

ll Scan() {
    int ch, flag = 0;
    ll res = 0;
    if((ch = getchar()) == '-')
        flag = 1;
    else if(ch >= '0' && ch <= '9')
        res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9' )
        res = res * 10 + ch - '0';
    return flag ? -res : res;
}

BigNum a, b, c, d;

int main() {
    ll t, n;
    t = Scan();
    for (ll cas = 1; cas <= t; cas++) {
        a = BigNum(Scan());
        b = BigNum(8);
        c = BigNum(7);
        d = BigNum(1);
        a = a * a * b - a * c + d;
        printf("Case #%lld: ", cas);
        a.print();
    }
    return 0;
}