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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 ?
● 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)
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：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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
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<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;
}
BigNum BigNum::operator^(const int &n)const //大数的n次方运算
{
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");
}
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;
}HDU - 5047 Sawtooth
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