openjudge divisibility_consider an arbitrary sequence of integers. one ca-优快云博客

本文链接：https://blog.youkuaiyun.com/clover_hxy/article/details/50151367

本文探讨了如何通过在一组整数序列中插入加号或减号，来判断是否存在运算组合使得最终结果能被特定整数K整除。详细介绍了输入输出格式、动态规划算法的应用，并提供了示例输入输出以清晰阐述解决问题的方法。

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747:Divisibility

描述

Consider an arbitrary sequence of integers. One can place + or - operatorsbetween integers in the sequence, thus deriving different arithmeticalexpressions that evaluate to different values. Let us, for example, take thesequence: 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 beplaced between integers in the sequence in such way that resulting value isdivisible 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 ofintegers.

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. Eachinteger is not greater than 10000 by it's absolute value.

输出

Write to the output file the word "Divisible" if given sequenceof 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

题目大意：在一串数中插于加号或减号，使最后的结果是k的倍数，如果可以输出Divisible，否则输出Not divisible。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,i,j,k;
int num[100100],now[210],last[210];//last数组存储的是上一个数计算完后的状态，然后在此基础上计算即可
int main()
{
scanf("%d%d",&n,&k);
for (i=1;i<=n;i++)
  scanf("%d",&num[i]);
now[100]=1;
for (i=1;i<=n;i++)
  {
    num[i]%=k;
    for (j=100-k;j<=100+k;j++)
      {
          last[j]=now[j];
          now[j]=0;
      }
    for (j=100-k;j<=100+k;j++)//动态规划部分
      if (last[j]==1)
       {
          now[(j-100-num[i])%k+100]=1;
          now[(j-100+num[i])%k+100]=1;
       }
  }
if (now[100]==1)
cout<<"Divisible";
else
  cout<<"Notdivisible";
return 0;
}