HDU1753

話說，經過了漫長的一個多月，小明已經成長了許多，所以他改了一個名字叫“大明”。?
這時他已經不是那個只會做100以內加法的那個“小明”了，現在他甚至會任意長度的正小數的加法。?

現在，給你兩個正的小數A和B，你的任務是代表大明計算出A+B的值。?

Input

本題目包含多組測試數據，請處理到文件結束。?
每一組測試數據在一行里面包含兩個長度不大于400的正小數A和B。

Output

請在一行里面輸出輸出A+B的值，請輸出最簡形式。詳細要求請見Sample Output。?

Sample Input

1.1 2.9 1.1111111111 2.3444323343 1 1.1

Sample Output

4 3.4555434454 2.1 import java.util.*; import java.math.*; public class Main {static int maxn=105;static int mod=(int)(1e9+7);public static void main(String[] args) {Scanner cin=new Scanner(System.in);int ca=0;while(cin.hasNext()) {ca++;BigDecimal a=cin.nextBigDecimal();BigDecimal b=cin.nextBigDecimal();System.out.println(a.add(b).stripTrailingZeros().toPlainString());//先去掉末尾的后導0，然后轉化為字符串}cin.close();} }

51Nod1030

給出一個36進制的大數(0-9,A-Z)，將其轉為10進制并輸出。

Input

輸入：36進制的大數，每一位用0-9,A-Z來表示，A表示10，Z表示35。(A的長度 <= 100000)

Output

輸出：該數的10進制表示

Sample Input

1AZ

Sample Output

1691 import java.math.*; import java.util.*; public class Main { //num需要轉換的數字，from原數的進制，to要轉換的進制private static String change(String num,int from,int to) {return new java.math.BigInteger(num,from).toString(to);}public static void main (String[] args) {Scanner cin = new Scanner(System.in);while(cin.hasNext()) {String num=cin.next();System.out.println(change(num,36,10));}cin.close();}}

超時代碼

import java.math.*;import java.util.*; public class Main {void solve () {Scanner cin = new Scanner(System.in);int maxn=(int)(1e5+6); //int T=cin.nextInt();int ca=0;while(cin.hasNext()) {ca++; String num=cin.next();BigInteger ans=BigInteger.ZERO;int len=num.length();int k=num.length()-1;for(int i=0;i<len;i++) {char ch=num.charAt(i);int m=0;if(ch>='0'&&ch<='9')m=ch-'0';else if(ch>='A'&&ch<='Z')m=ch-'A'+10;BigInteger p=BigInteger.valueOf(36);p=p.pow(k);ans=ans.add(p.multiply(BigInteger.valueOf(m)));k--;}System.out.println(ans);}cin.close();}public static void main (String[] args) {Main work = new Main();work.solve ();} }

51Nod 1166

給出一個大整數N，求不大于N的平方根的最大整數。例如：N = 8，2 * 2 < 8，3 * 3 > 8，所以輸出2。

Input

給出一個大數N（N的長度 <= 100000)。

Output

輸出不大于Sqrt(n)的最大整數。

Sample Input

9

Sample Output

3 from math import * from decimal import * getcontext().prec=10**5 print(floor(Decimal((input())).sqrt())) import java.math.*; import java.util.*; public class Main {//牛頓迭代法void solve () {Scanner cin = new Scanner(System.in);while(cin.hasNext()) {String s=cin.next();int len=s.length();if(len<9) {System.out.println((long)Math.sqrt(Double.valueOf(s)));continue;}BigInteger num=new BigInteger(s);if(len%2==0) {s=s.substring(0, len/2+1);}else {s=s.substring(0,(len+1)/2);}BigInteger x=new BigInteger(s);if(s.equals("1"))System.out.println("1");else {while(num.compareTo(x.multiply(x))<0) {x=x.add(num.divide(x)).divide(new BigInteger("2"));}}System.out.println(x);}cin.close();}//num需要轉換的數字，from原數的進制，to要轉換的進制private static String change(String num,int from,int to) {return new java.math.BigInteger(num,from).toString(to);}public static void main (String[] args) {Main work = new Main();work.solve ();} }

HDU1865

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.?

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.?

Output

The output contain n lines, each line output the number of result you can get .?

Sample Input

3 1 11 11111

Sample Output

1 2 8 import java.math.*; import java.util.*; public class Main {void solve () {Scanner cin = new Scanner(System.in);int maxn=207;BigInteger[] f=new BigInteger[maxn];f[1]=BigInteger.valueOf(1);f[2]=BigInteger.valueOf(2);for(int i=3;i<maxn;i++) {f[i]=f[i-1].add(f[i-2]);}int T=cin.nextInt();while((T--)!=0) {String n=cin.next();int len=n.length();System.out.println(f[len].toString());}cin.close();}public static void main (String[] args) {Main work = new Main();work.solve ();} }

?

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