POJ1745Divisibility【dp】
来源:互联网 发布:网络拓扑发现算法java 编辑:程序博客网 时间:2024/06/03 15:56
Divisibility
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 11053
Accepted: 3953
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
4 717 5 -21 15
Sample Output
Divisible
题意:给出N个数通过在其中添加+-号问是否能计算出k的倍数
dp[i][j]代表前i个数余数为j是否存在
#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;const int MAX=10010;int dp[MAX][110];int num[MAX];int main(){int n,m,k,i,j;while(scanf("%d%d",&n,&k)!=EOF){memset(dp,0,sizeof(dp));for(i=0;i<n;++i){scanf("%d",&num[i]);if(num[i]<0){num[i]*=-1;}num[i]%=k;}dp[0][num[0]]=1;for(i=1;i<n;++i){for(j=0;j<k;++j){if(dp[i-1][j]){dp[i][(j+num[i])%k]=1;dp[i][(k+j-num[i])%k]=1;}}}if(dp[n-1][0])printf("Divisible\n");elseprintf("Not divisible\n");}return 0;}
- POJ1745Divisibility【dp】
- POJ1745Divisibility(dp)(AC)
- POJ1745Divisibility(dp)
- POJ1745Divisibility题解动态规划DP
- poj1745Divisibility
- POJ1745Divisibility
- dp
- dp
- dp
- 【DP】
- dp
- dp
- DP
- DP
- DP
- DP
- DP
- dp
- 已解决:Was启动报错 mbind:Invalid argument
- Android.mk 语法
- OO真经——关于面向对象的哲学体系及科学体系的探讨(上)
- 双机热备数据库备份脚本 v2.1
- Android—Handler.obtainMessage()与new Message()相比
- POJ1745Divisibility【dp】
- 十四、第三章再续:快速选择SELECT算法的深入分析与实现
- NSString [a compare:b] NSOrderedSame NSOrderedAscending NSOrderedDescending字符串比较
- arcLength函数
- Linux命令之——ps命令详解
- POJ2388map的应用统计树的百分比,并按照字母顺序输出
- OutMan——集合对象的内存管理、copy的介绍及使用
- 传值API Url 中有小数点 导致路径访问不到
- python cmp函数详解