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acm_icpc網(wǎng)絡(luò)賽第六站：上海賽區(qū)（跪爛心已死）

來源：程序員人生   發(fā)布時間：2014-10-11 08:00:01 閱讀次數(shù)：3896次

最后一站了，很無奈的是還是沒出線，最終3題結(jié)束，排名380 不得不承認自己不行，總覺著自己才弄了幾個月，然后各種為自己的弱找借口，不行就是不行，沒有借口，以后要更加努力了。廢話不多說了 Fighting！

1006：

Sawtooth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 466    Accepted Submission(s): 152


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
遞推+大數(shù)
增加第n+1個 “M”時，交點最多有16n個，所以增加區(qū)域 16n+1個；
f(n+1)-f(n)=16*n+1;
f(1)=2;
解得
f(n)=n*(8*n-7)+1;
然后 直接上BigInteger 就行了 然后悲劇的是當時我不會java的輸入優(yōu)化結(jié)果出題人故意卡java。。最后隊友敲的數(shù)組模擬過的
下面上java輸入輸出優(yōu)化的代碼 很簡單~
import java.io.*;
import java.util.*;
import java.math.BigInteger;
public class Hello {
	public static void main(String[] args) throws IOException{
		StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t,cas=1;BigInteger n,a,b,c;
        in.nextToken();
        t=(int)in.nval;
        for(int i=1;i<=t;i++){
        	in.nextToken();
        	long x=(long)in.nval;
        	n=BigInteger.valueOf(x);
        	a=new BigInteger("8");
        	b=new BigInteger("7");
        	c=new BigInteger("1");
        	BigInteger ans=a.multiply(n);
        	ans=ans.subtract(b);
        	ans=ans.multiply(n);
        	ans=ans.add(c);
        	out.println("Case #"+cas+": "+ans);cas++;
        }
        out.flush();//刷新緩沖區(qū) （必須）
        out.close();
	}
}

1009：

Divided Land

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 133    Accepted Submission(s): 69


Problem Description
It’s time to fight the local despots and redistribute the land. There is a rectangular piece of land granted from the government, whose length and width are both in binary form. As the mayor, you must segment the land into multiple squares of equal size for the villagers. What are required is there must be no any waste and each single segmented square land has as large area as possible. The width of the segmented square land is also binary.
 

Input
The first line of the input is T (1 ≤ T ≤ 100), which stands for the number of test cases you need to solve.

Each case contains two binary number represents the length L and the width W of given land. (0 < L, W ≤ 21000)
 

Output
For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then one number means the largest width of land that can be divided from input data. And it will be show in binary. Do not have any useless number or space.
 

Sample Input
3 10 100 100 110 10010 1100
 

Sample Output
Case #1: 10 Case #2: 10 Case #3: 110
二進制下求a,b的最大公約數(shù) 還是BigInteger 由于一開始不知道BigInteger二進制的轉(zhuǎn)換 chp用模擬敲了半天也沒搞好 后來用java 10幾行解決了
import java.io.*;
import java.util.*;
import java.math.*;
import java.text.*;
public class Main {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        String a,b;int t,cas=1;
        t=in.nextInt();
        for(int i=0;i<t;i++){
            a=in.next();
            b=in.next();
            BigInteger x=new BigInteger(a,2);
            BigInteger y=new BigInteger(b,2);
            System.out.print("Case #"+cas+": ");cas++;
            System.out.println(x.gcd(y).toString(2));
        }
        
    }
}

1012：

the Sum of Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 181    Accepted Submission(s): 114


Problem Description
A range is given, the begin and the end are both integers. You should sum the cube of all the integers in the range.
 

Input
The first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve.
Each case of input is a pair of integer A,B(0 < A <= B <= 10000),representing the range[A,B].
 

Output
For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then output the answer
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