HDU 5047 Sawtooth(有趣的思维题+证明)

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;
}
时间: 2024-10-07 06:14:45

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