HDU 5047 Sawtooth （JAVA大數(shù)類）
時間：2020-01-08 13:40:01
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[導讀]題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5047
題面：
Sawtooth
Time Limit: 2000/1000 MS (Java/Ot
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5047
題面：
Sawtooth Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1636????Accepted Submission(s): 637
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 ?
Source 2014 ACM/ICPC Asia Regional Shanghai Online ? 解題：
??? 每一個M四條邊，每條邊最多與原來的邊相交，形成原來的邊數(shù)*4個交點，就多出來相應這么多的面，因為原來M內部也有一個面，故還需加一。
??? 所以，F(xiàn)（n）=F（n-1）+4*4*（n-1）+1。不斷迭代可以將這條公式化成F（n）=8*n*n-7*n+1。因為數(shù)據(jù)范圍很大，所以要用大數(shù)。
??? 直接用公式配合C++大數(shù)模板T了。于是，我用JAVA大數(shù)類交了一發(fā),1900ms，不得不說時間卡得真緊。上網一查，原來JAVA的讀入輸出是能夠優(yōu)化的，詳見這篇博客。我自己用大數(shù)交了一下800ms，不過貌似也可以在大數(shù)乘上優(yōu)化。
代碼（JAVA）：
import java.io.*;
import java.util.*;
import java.math.*;
;public class Main {
public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
int t=cin.nextInt();
for(int i=1;i<=t;i=i+1)
{
BigInteger ans,tmp;
BigInteger n=cin.nextBigInteger();
ans=n.multiply(n);
ans=ans.multiply(BigInteger.valueOf(8));
tmp=n.multiply(BigInteger.valueOf(7));
ans=ans.subtract(tmp);
ans=ans.add(BigInteger.ONE);
System.out.println("Case #"+i+": "+ans);
}
}
}
代碼（JAVA優(yōu)化版）：
import java.io.*;
import java.util.*;
import java.math.*;
;public class Main {
public static void main(String args[])
{
Scanner sc = new Scanner(new BufferedInputStream(System.in));
PrintWriter cout=new PrintWriter(System.out);
int t=sc.nextInt();
for(int i=1;i<=t;i=i+1)
{
BigInteger ans,tmp;
BigInteger n=sc.nextBigInteger();
ans=n.multiply(n);
ans=ans.multiply(BigInteger.valueOf(8));
tmp=n.multiply(BigInteger.valueOf(7));
ans=ans.subtract(tmp);
ans=ans.add(BigInteger.ONE);
cout.println("Case #"+i+": "+ans);
}
cout.flush();
}
}代碼（C++大數(shù)模板）：
#include
#include
#include
#include
#include
using namespace std;
#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4
class BigNum
{
private:
int a[500]; //可以控制大數(shù)的位數(shù)
int len; //大數(shù)長度
public:
BigNum(){ len = 1;memset(a,0,sizeof(a)); } //構造函數(shù)
BigNum(const int); //將一個int類型的變量轉化為大數(shù)
BigNum(const char*); //將一個字符串類型的變量轉化為大數(shù)
BigNum(const BigNum &); //拷貝構造函數(shù)
BigNum &operator=(const BigNum &); //重載賦值運算符，大數(shù)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(shù)之間的相加運算
BigNum operator-(const BigNum &) const; //重載減法運算符，兩個大數(shù)之間的相減運算
BigNum operator*(const BigNum &) const; //重載乘法運算符，兩個大數(shù)之間的相乘運算
BigNum operator/(const int &) const; //重載除法運算符，大數(shù)對一個整數(shù)進行相除運算
BigNum operator^(const int &) const; //大數(shù)的n次方運算
int operator%(const int &) const; //大數(shù)對一個int類型的變量進行取模運算
bool operator>(const BigNum & T)const; //大數(shù)和另一個大數(shù)的大小比較
bool operator>(const int & t)const; //大數(shù)和一個int類型的變量的大小比較
void print(); //輸出大數(shù)
};
BigNum::BigNum(const int b) //將一個int類型的變量轉化為大數(shù)
{
int 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(shù)
{
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(shù)
{
int i;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n) //重載賦值運算符，大數(shù)之間進行賦值運算
{
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) //重載輸入運算符
{
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 << b.a[b.len - 1];
for(i = b.len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << b.a[i];
}
return out;
}
BigNum BigNum::operator+(const BigNum & T) const //兩個大數(shù)之間的相加運算
{
BigNum t(*this);
int i,big; //位數(shù)
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 //兩個大數(shù)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 ; 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[t1.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(shù)之間的相乘運算
{
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 //大數(shù)對一個整數(shù)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(shù)對一個int類型的變量進行取模運算
{
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(shù)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(shù)和另一個大數(shù)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(shù)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(shù)
{
int i;
cout << a[len - 1];
for(i = len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << a[i];
}
cout << endl;
}
int main(void)
{
int t;
BigNum n,ans;
cin>>t;
for(int i=1;i<=t;i++)
{
cin>>n;
cout<<"Case #"< 
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