描述
Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are eight possible expressions: 17 + 5 + -21 + 15 = 16
17 + 5 + -21 - 15 = -14
17 + 5 - -21 + 15 = 58
17 + 5 - -21 - 15 = 28
17 - 5 + -21 + 15 = 6
17 - 5 + -21 - 15 = -24
17 - 5 - -21 + 15 = 48
17 - 5 - -21 - 15 = 18
We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5+-21-15=-14) but is not divisible by 5.
You are to write a program that will determine divisibility of sequence of integers.
输入
The first line of the input file contains two integers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by a space.
The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it's absolute value.
输出
Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.
样例输入
4 7
17 5 -21 15
样例输出
Divisible
来源
Northeastern Europe 1999
1 #include <cstdio> 2 #include <cstring> 3 #include <memory.h> 4 #include <algorithm> 5 #include <stdlib.h> 6 #include <math.h> 7 #include <iostream> 8 #include<queue> 9 #include <vector> 10 #include <bitset> 11 using namespace std; 12 13 int dp[2][105]; 14 int n, k; 15 int num[10005]; 16 17 void solve() { 18 dp[1][num[1]] = 1; 19 for (int i = 2; i <= n; i++) 20 { 21 for (int j = 0; j < k; j++)dp[i % 2][j] = 0; 22 for (int j = 0; j < k; j++) 23 if (dp[(i - 1) % 2][j]) 24 { 25 dp[i % 2][(num[i] + j) % k] = 1; 26 dp[i % 2][(k+j - num[i]) % k] = 1; 27 } 28 } 29 if (dp[n%2][0]) 30 printf("Divisible "); 31 else 32 printf("Not divisible "); 33 } 34 35 int main() 36 { 37 scanf("%d%d", &n, &k); 38 for (int i = 1; i <= n; i++) 39 { 40 scanf("%d", &num[i]); 41 num[i] = abs(num[i]) % k; 42 } 43 memset(dp, 0, sizeof(dp)); 44 solve(); 45 return 0; 46 }
想用滚动数组结果忘了每次初始化，debug很久