Java 大数加减乘除 hdu 5047

acm.hdu.edu.cn/showproblem.php?pid=5047

Sawtooth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1592    Accepted Submission(s): 622

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

212

 

Sample Output

Case #1: 2Case #2: 19

import java.io.*;import java.math.*;public class Main{  public static BufferedReader cin=new BufferedReader(new InputStreamReader(System.in));  public static BufferedWriter cout=new BufferedWriter(new OutputStreamWriter(System.out));  public static void main(String []args) throws Exception{    int T=Integer.parseInt(cin.readLine());//读取T组数据    for(int nkase=1;nkase<=T;nkase++){      cout.write("Case #"+nkase+": ");      BigInteger N=new BigInteger(cin.readLine());//输入      BigInteger ans=N.multiply/*乘法*/(N).multiply(BigInteger.valueOf(8)/*将int的  8转化为biginteger*/).subtract/*减法*/(N.multiply(BigInteger.valueOf(7))).add(BigInteger.valueOf(1));      cout.write(ans.toString());//转化成字符串      cout.newLine();//换行    }    cout.flush();    cout.close();//关闭  }}、、、、//m.divide(n)===m/n