[Huawei machine trial exercise questions] 14. Separated by integers, Huawei exercises

[Huawei machine trial exercise questions] 14. Separated by integers, Huawei exercises

Question

Description: an integer can be divided into the sum of the power of 2. For example: 7 = 1 + 2 + 47 = 1 + 2 + 2 + 27 = 1 + 1 + 1 + 47 = 1 + 1 + 1 + 2 + 27 = 1 + 1 + 1 + 1 + 1 + 1 + 27 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 there are six different splitting methods. For example, 4 can be split into 4 = 1 + 1 + 1 + = 2 + = 1 + 1 + 2. F (n) indicates the number of different splits of n, for example, f (7) = 6. write a program, read n (no more than 1000000), and Output f (n) % 1000000000. Question type: null difficulty: Primary running time limit: 10Sec memory limit: 128MByte stage: pre-entry exercise input: each group of inputs contains an integer: N (1 <= N <= 1000000 ). Output: f (n) % 1000000000 is output for each group of data. The output contains multiple rows, with one result per row. If the input data is out of the range, output-1. Sample input: 7 sample output: 6

Ideas

If n = 2 k + 1 is an odd number, f (2 k + 1) = f (2 k ). In fact, the first item of the 2 k + 1 split must be 1. If this 1 item is removed, it will be the same as the 2 k split. When n = 2 k is an even number, we consider splitting between 1 and 1. If one exists, the first two items are 1, which is the same as the 2k-2 split. If there is no 1, divide each item by 2, and the split is the same as that of k. Therefore, f (2 k) = f (2k-2) + f (k );

Code 1

/* ------------------------------------- * Date: 2015-06-30 * Author: SJF0115 * Subject: integer separation * Source: huawei machine --------------------------------------- */# include <iostream> # include <string> # include <algorithm> # include <vector> using namespace std; # define Max limit 00int Count [Max + 1]; int Split (int n) {Count [1] = 1; Count [2] = 2; for (int I = 3; I <= n; ++ I) {// odd if (I % 2) {Count [I] = Count [I-1];} // if else {Count [I] = (Count [I-2] + Count [I/2]) % 1000000000 ;} // else} // for return Count [n];} int main () {int n; // freopen ("C: \ Users \ Administrator \ Desktop \ c00000000.txt "," r ", stdin); while (cin> n) {if (n> Max | n <0) {return-1;} // if cout <Split (n) <endl ;}// while return 0 ;}

Code 2

Timeout

/* ------------------------------------- * Date: 2015-06-30 * Author: SJF0115 * Subject: integer separation * Source: huawei machine --------------------------------------- */# include <iostream> # include <string> # include <algorithm> # include <vector> using namespace std; # define Max limit 00int Split (int n) {if (n = 1) {return 1;} // if (n = 2) {return 2;} // if // odd if (n % 2) {return Split (n-1) % 1000000000;} // if else {return Split (n-2) % 1000000000 + Split (n/2) % 1000000000 ;} // else} int main () {int n; // freopen ("C: \ Users \ Administrator \ Desktop \ c00000000.txt", "r", stdin ); while (cin >>n) {if (n> Max) {return-1 ;}// if cout <Split (n) <endl ;} // while return 0 ;}

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