Description
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.Input
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.Output
Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.Sample Input
4 7 17 5 -21 15Sample Output
Divisible题意：输入n个数，通过添加+和-能否是的结果对k取余为0思路:智商再次背碾压首先一个数，不用说，第一个数之前不用加符号就是本身，那么本身直接对K取余，那么取17的时候有个余数为2然后来了一个5，(2 + 5)对7取余为0(2 - 5)对7取余为4（将取余的负数变正）那么前2个数有余数0和4再来一个-21（0+21）对7取余为0（0-21）对7取余为0（4+21）对7取余为4（4-21）对7取余为4再来一个-15同样是这样（0+15）%7 = 1（0-15）%7 = 6（4+15）%7 = 5（4-15）%7 = 3同理可以找到规律，定义dp[i][j]为前i个数进来余数等于j是不是成立，1为成立，0为不成立那么如果dp[N][0]为1那么即可以组成一个数对K取余为0初始化dp为0
然后dp[1][a[1]%k] = 1for i = 2 to N dofor j = 0 to K do if(dp[i - 1][j]) dp[i][(j + a[i])%k] = 1; dp[i][(j - a[i])%k] = 1; if endfor endfor end
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 int dp[10000 + 10][100 + 10],a[10000 + 10]; 7 int n,k; 8 int mod(int x) 9 { 10 if(x < 0) 11 { 12 return x + k; 13 } 14 else 15 return x; 16 } 17 int main() 18 { 19 while(scanf("%d%d", &n, &k) != EOF) 20 { 21 for(int i = 1; i <= n; i++) 22 scanf("%d", &a[i]); 23 memset(dp, 0, sizeof(dp)); 24 dp[1][ mod(a[1] % k) ] = 1; 25 for(int i = 2; i <= n; i++) 26 { 27 for(int j = 0; j <= k; j++) 28 { 29 if(dp[i - 1][j]) 30 { 31 dp[i][ mod((j + a[i]) % k)] = 1; 32 dp[i][ mod((j - a[i]) % k)] = 1; 33 } 34 } 35 } 36 if(dp[n][0]) 37 printf("Divisible\n"); 38 else 39 printf("Not divisible\n"); 40 } 41 42 43 return 0; 44 } View Code
转载于:https://www.cnblogs.com/zhaopAC/p/5055085.html
