HDU 5047 Sawtooth （JAVA大數類）

標籤：

題目連結：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 DescriptionThink 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 ?

 
InputThe 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) 
OutputFor 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

212

 
Sample Output

Case #1: 2Case #2: 19

 
Source2014 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4*（n-1）個交點，是不是和歐拉定理有關？？然後隊友就神奇搬地推出f(n)=f(n-1)+4*4*(n-1)+1;然後就出來了8*n^2-7*n+1...

    好吧，重點是他直接用公式配合C++大數模板T了。於是，我用JAVA大數類交了一發,1900ms，不得不說時間卡得真緊。上網一查，原來JAVA的讀入輸出是能夠最佳化的，詳見這篇部落格。我自己用大數交了一下800ms，不過貌似也可以在大數乘上最佳化。

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最佳化版）：

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++大數模板）：

#include<iostream> #include<string> #include<iomanip> #include<cstring>#include<algorithm> using namespace std; #define MAXN 9999#define MAXSIZE 10#define DLEN 4class BigNum{ private: int a[500];    //可以控制大數的位元 int len;       //大數長度public: BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //建構函式BigNum(const int);       //將一個int類型的變數轉化為大數BigNum(const char*);     //將一個字串類型的變數轉化為大數BigNum(const BigNum &);  //拷貝建構函式BigNum &operator=(const BigNum &);   //重載賦值運算子，大數之間進行賦值運算friend istream& operator>>(istream&,  BigNum&);   //重載輸入運算子friend ostream& operator<<(ostream&,  BigNum&);   //重載輸出運算子BigNum operator+(const BigNum &) const;   //重載加法運算子，兩個大數之間的相加運算 BigNum operator-(const BigNum &) const;   //重載減法運算子，兩個大數之間的相減運算 BigNum operator*(const BigNum &) const;   //重載乘法運算子，兩個大數之間的相乘運算 BigNum operator/(const int   &) const;    //重載除法運算子，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類型的變數進行模數運算    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類型的變數的大小比較void print();       //輸出大數}; BigNum::BigNum(const int b)     //將一個int類型的變數轉化為大數{ 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)     //將一個字串類型的變數轉化為大數{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)  //拷貝建構函式{ int i; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++)a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n)   //重載賦值運算子，大數之間進行賦值運算{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)   //重載輸入運算子{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)   //重載輸出運算子{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   //兩個大數之間的相加運算{BigNum t(*this);int i,big;      //位元   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; elset.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 ; 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]; } elset1.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類型的變數進行模數運算    {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; elsereturn false; } elsereturn false; }bool BigNum::operator >(const int & t) const    //大數和一個int類型的變數的大小比較{BigNum b(t);return *this>b;}void BigNum::print()    //輸出大數{ 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 #"<<i<<": ";ans=BigNum(8)*n*n-BigNum(7)*n+BigNum(1);cout<<ans<<endl;}return 0;}

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HDU 5047 Sawtooth （JAVA大數類）