题目来源：http://poj.org/problem?id=1745
Divisibility
Time Limit: 1000MS |
Memory Limit: 10000K |
Description
Consider an arbitrary sequence of integers. Onecan place + or - operators between integers in the sequence, thus derivingdifferent arithmetical expressions that evaluate to different values. Let us,for example, take the sequence: 17, 5, -21, 15. There are eight possibleexpressions: 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.
Input
The first line of the input file contains twointegers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by aspace.
The second line contains a sequence of N integers separated by spaces. Eachinteger 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 15
SampleOutput
Divisible
Source
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解题思路
动态规划（0-1背包）
题目本身的坑点在于C/C++里 -20%7 = -6而不是1，需要+7化成正数
值得注意的是之前一直听说但从来没有遇到过的”cin vs. scanf””int a[10001] vs. int *a = new int*[n]”前者速度慢于后者的情况终于发生了。
百练上的数据比较弱，用后面的写法也能过。poj上的数据比较严，一定要直接开固定大小的数组才能过，动态分配数组过不了；用cin也能过，但cin版本用时797ms, scanf版本用时297ms，差了500ms！太可怕了！
血泪教训：
1. 如果题目告诉了数据个数的上限，直接按上限开固定大小的数组；
2. 如果要从标准输入流中读取大量数据，用scanf不要用cin.
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代码
//E:Divisibility
//总时间限制: 1000ms 内存限制: 65536kB
//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
#include<fstream>
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
#ifndef ONLINE_JUDGE
ifstream fin("tm201602E.txt");
int n,k,i,j;
fin >> n >> k;
int *num = new int[n];
for (i=0; i<n; i++)
{
fin >> num[i];
num[i] %= k;
if (num[i]<0) // 防止出现负数
{
num[i] += k;
}
}
int **dp = new int*[n];
for (i=0; i<n; i++)
{
dp[i] = new int[k]();
}
// dynamic programming
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][(j-num[i]+k)%k] = 1; // 防止出现负数
}
}
}
if (dp[n-1][0] == 1)
{
cout << "Divisible";
}
else
{
cout << "Not divisible";
}
delete[] num;
for (i=0; i<n; i++)
{
delete[] dp[i];
}
delete[] dp;
fin.close();
#endif
#ifdef ONLINE_JUDGE
int n,k,i,j,tmp;
cin >> n >> k;
int num[10001]; // 开固定大小数组
for (i=0; i<n; i++)
{
cin >> num[i]; // 这里用cin也能过，但最好用scanf
num[i] %= k;
if (num[i]<0)
{
num[i] += k;
}
}
int dp[10001][100] = {} ;
// dynamic programming
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][(j-num[i]+k)%k] = 1;
}
}
}
if (dp[n-1][0] == 1)
{
cout << "Divisible";
}
else
{
cout << "Not divisible";
}
#endif
}