Casual association of a Baidu star Programming Competition question · (2)

Baidu star is the world's largest Chinese search engine. Baidu holds a high-level Program Design Competition for Chinese college students and programmers. All his questions are algorithm questions.

Although I am a. Net programmer, I love algorithms after work. I think this question is a bit interesting, so I will share it with you. I think of two ways to provide you with an open mind.

Solution 2 is introduced below.

Solution 2: Small-sized amplification. Small and large. First, let's talk about my analysis ideas.

The first step is to judge whether I is no less than I/2 and then use this loop.

Step 2: Determine whether the value accumulation in this range is equal to a specific value.

Step 3: Output

Step 4: output the value of none if no value exists.

Now let's look at the corresponding source code.

1 Console. writeline ("enter a positive integer"); 2 int inputnum = convert. toint32 (console. readline (); 3 // indicates whether consecutive values are identified; 4 bool flag = false; 5 // the corresponding counter variable 6 int I = 1, j = 2, sum = 0; 7 // 8 while (I <inputnum/2) 9 {10 sum = add (I, j); 11 while (sum! = Inputnum & I <inputnum/2 + 1) 12 {13 if (sum> inputnum) 14 I ++; 15 else16 J ++; 17 sum = add (I, j); 18} 19 // print the corresponding loop 20 for (int K = I; k <= J & J> I; k ++) 21 {22 23 console. write (k); 24 25 console. write (""); 26} 27 I ++; 28 console. writeline (); 29} 30 31 if (! Flag) 32 {33 console. writeline ("NONE"); 34} 35 console. readkey (); 36 // calculate the accumulated algorithm 37 static int add (int m, int N) 38 {39 int sum = 0; 40 for (INT I = m; I <= N; I ++) 41 {42 sum + = I; 43} 44 return sum; 45}

The function of this algorithm is to see whether it is equal through loops.

He is different from the first algorithm. The first method is to subtract it from it to see if it can get 0. The second method is used to accumulate the algorithm to see if it is equal. And there are two pointer variables scanning each other.

All in all, these two algorithms have their own merits.