java power of 2

# Include Using namespace STD; // The first method exploits the loopBool makedemo-( const Int M, Int POW) // POW indicates how many power m is 2{Int I = 2;Int S = 1;Pow = 0;If (M If (M = 0) return true;Do{S * = I;Pow ++;} While (S If (S = m) return true;Else{Pow = 0;Return false;}} // The second method. If M/2

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Topic Connection: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1048 ————————————————————————————————————-.Integer decomposition to a power of 2 Base time limit: 3 seconds Space limit: 131072 KB score: 1280 Difficulty: 9-level algorithm problem Any positive integer can be decomposed into a power of 2, g

// Determine whether N is a power of 2 # include

Turn | How to quickly determine whether a positive integer is a power of 2Turn from past memory: http://www.iteblog.com/archives/716The power of 2 has the following two characteristics:1.20=1, 0001,21=2, 0010,22=4, 0100,23=8, 10,002, there is only one 1, and 1 followed by N 0.2. If the number minus 1 will be found that

B. jzzhu and sequencestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output Jzzhu has sorted Ted a kind of sequences, they meet the following property: You are givenXAndY, Please calculateFNModulo 1000000007 (109? +? 7 ).Input The first line contains two integersXAndY(|X| ,? |Y|? ≤? (109). The second line contains a single integerN(1? ≤?N? ≤? 2 · 109 ).Output Output a single integer representingFNModulo

Judge whether a value is a power of 2. Common functions in 3D games In the past, the judgment method was to cyclically determine bits or directly divide them continuously. The efficiency of these methods can be imagined. Here I use a direct bit operation to get the efficiency of the Code. first look at the Code: Bool Is2Power (int n) { If (n = 1) Return false; If ((((~ N) (n-1) + 1) = n) Return t

This is an interview question. First, we can analyze the number of power forms of 2. the binary form must be 1 bits and the rest bits must be 0 Therefore, an intuitive method is to compare it with the N power of all 2, because the unsigned integer we usually consider is 32-bit, therefore, there are 32

Ispoweroftwo determines the power of 2, and ispoweroftwo determines The first result is: Public bool IsPowerOfTwo (int n ){If (n Return (n (n-1) = 0 );} Analysis: The power of 2 can be calculated using the Left shift ( N (n-1): Change the bitwise of the binary representation of n to 0. Let's look at a simple example:

AbstractY = a + B; how can we use digital reality for a very simple computing? IntroductionUse environment: US us II 8.0 Y = a + B; this is a simple operation in C. However, if you want to use OpenGL on the digital path, beginners may encounter some difficulties. Y = a + B Do not use clockAdd2_assign.v/OpenGL Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> 1 Module Add2_assign ( 2

Dating with Girls (2) Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2659 Accepted Submission (s): 744 Problem Description If You had solved the problem Dating with girls (1). I Think you can solve this problem too. This problem was also about dating with girls. Now you are in a maze and the girl you want t

Question Link This question seems like the basic idea of POJ3233 is the same. This question should be obtained using a fast power. If there are many proportional series of items, Division numbers should be included in the summation formula. Therefore, mod cannot be taken directly, and fast rice conversion should be performed. For example, sum = 2 ^ 1 + 2 ^

Recently see the source code of the Concurrenthashmap class in the JDK, which has a function:/*** Returns a power of both table size for the given desired capacity. * See Hackers Delight, SEC 3.2* /private static final int tablesizefor (int c) { int n = c-1; n |= N >>> ; 1 n|= n >>> 2 n|= n >>> 4 n|= n >>> 8 n|= n >>> return (n ;} I c

(); + } A while('0''9') at { -ans=ans*Ten+s-'0'; -s=GetChar (); - } - returnans*ff; - } in Long LongQuick_mod (ll A,ll b)/*slow multiplication prevents overflow*/ - { toa%=q;b%=Q; +ll ans=0; - while(b) the { * if(b1) $ {Panax Notoginsengans+=A; -ans%=q;// the } +b>>=1; Aa1; thea%=Q; + } - returnans; $ } $ Jz martax (Jz x,jz y) - { - Jz ans; theAns.line=X.line; -Ans.cal=y.cal;Wuyi for(intI=1; ii) the for(intj=1; jj) -

1 PackageCom.java7;2 3 Public classWhiledemo {4 Public Static voidMain (string[] args) {5 inte;6 intresult;7 for(inti = 0; I ){8result = 1;9E =i;Ten while(E > 0){ OneResult *= 2; Ae--; - } - theSystem.out.println ("2 to the" + i + "power is" +result); - } - } -}Execution process:1:i

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