1745: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

我自己做的时候因为n范围太大，超时了，我是直接枚举了，下面是错误代码：

#include<iostream>using namespace std;int n,k,a[11000],f[11000],res;bool g(){    res=a[0];    for(int i=1;i<n;i++){        if(f[i]==0){            res+=a[i];        }        if(f[i]==1){            res-=a[i];        }    }    //cout<<res<<endl;    if(res%k==0)return true;    else return false;}int main(){    cin>>n>>k;    for(int i=0;i<n;i++){        cin>>a[i];    }    while(f[n]!=1&&!g()){        f[1]++;        int c=1;        while(f[c]>1){            f[c]=0;            c++;            f[c]++;        }    }    if(f[n]==1){        cout<<"Not divisible"<<endl;    }    else cout<<"Divisible"<<endl;}

这个是参考的：状态：dp[i][j] 表示前 i 个数合成模 K 为 j 是否可行
状态转移方程：
dp[i + 1][((j + a[i + 1]) % K + K) % K] = 1;
dp[i + 1][((j - a[i + 1]) % K + K) % K] = 1;

#include <cstdio>#include <cstring>const int MAXN = 10000 + 10;const int MAXK = 100 + 10;int N, K;int a[MAXN];char dp[MAXN][MAXK];void Read(){    for (int i = 1; i <= N; ++i)    {        scanf("%d", a + i);    }}void Dp(){    memset(dp, 0, sizeof(dp));    dp[1][(a[1] % K + K) % K] = 1;    for (int i = 1; i < N; ++i)    {        for (int j = 0; j < K; ++j)        {            if (dp[i][j] != 0)            {                dp[i + 1][((j + a[i + 1]) % K + K) % K] = 1;                dp[i + 1][((j - a[i + 1]) % K + K) % K] = 1;            }        }    }    dp[N][0] == 0 ? puts("Not divisible") : puts("Divisible");}int main(){    while (scanf("%d%d", &N, &K) == 2)    {        Read();        Dp();    }    return 0;}