zoukankan      html  css  js  c++  java
  • UVA 10036 Divisibility

    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 a integer M indicating the number of cases to be analyzed. Then M couples of lines follow. 
    For each one of this couples, the first line 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

    For each case in the input file, 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

    2
    4 7
    17 5 -21 15
    4 5
    17 5 -21 15
    

    Sample Output

    Divisible
    Not divisible

    DP
    •f[i][j]=1表示前i个数字形成的表达式的值除以K之后可以余j
    •f[0][0]=1
    •考虑第i个数字x,假设它是正的。
    •如果 f[i-1][j] = 1,说明前i-1个数字的表达式的值除以K可以余j
    •在x的前面放“+”, f[i][(j+x)%K] = 1
    •在x的前面放“-”,f[i][(j-x+K)%K]=1
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 using namespace std;
     5 
     6 int N,K;
     7 bool f[2][100];
     8 
     9 int main()
    10 {
    11     int T ;
    12     scanf("%d",&T);
    13     while(T--) {
    14         scanf("%d%d" , &N ,&K) ;
    15         memset(f , false , sizeof(f)) ;
    16         f[0][0] = true ;
    17         int cur = 0 , x ;
    18         for(int i=1; i<=N; i++){
    19             cur ^= 1 ;
    20             memset(f[cur] , false , sizeof(f[cur])) ;
    21             scanf("%d" , &x);
    22             if(i>1 && x<0) x = -x;
    23             x%=K;
    24             for(int u = 0; u<K; u++) if(f[cur^1][u]) {
    25                 f[cur][(u+x+K)%K] = true;
    26                 if(i>1) f[cur][(u-x+K)%K]=true;
    27             }
    28         }
    29         if(f[cur][0]) puts("Divisible") ;
    30         else puts("Not divisible") ;
    31     }
    32     return 0;
    33 }
  • 相关阅读:
    JavaScript相关,this 的指向
    酷炫的渲染,CSS3动画!
    正襟危坐,聊一聊Web语义化
    本地存储的技术,localStorage、sessionStorage、cookie、session
    Express 系列(终章):MongoDB使数据持久化
    股票投资相关,成交量怎么看
    【日本語新聞選読】第5回:3月31日
    【1901日語听解2】第5回:3月30日
    【日語視聴説2】第5回:3月30日
    【日本語新聞編集】第4回:3月27日
  • 原文地址:https://www.cnblogs.com/zzy9669/p/3855159.html
Copyright © 2011-2022 走看看