HDOJ-5446 Clarke and problem(DP)

设d(i,j)d(i, j)d(i,j)表示前iii个数，模ppp为jjj的方案数，则容易得到d(0,0)=1,d(i,j)=d(i−1,j)+∑j=0p−1d(i−1,(j−a[i]) mod p)d(0,
0)=1, d(i, j)=d(i-1, j)+\sum_{j=0}^{p-1} d(i-1, (j-a[i]) \ mod \ p)d(0,0)=1,d(i,j)=d(i−1,j)+∑​j=0​p−1​​d(i−1,(j−a[i]) mod p)，很多人没1a是因为没注意∣ai∣≤109|a_i|
\le 10^9∣a​i​​∣≤10​9​​

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int N=1005;
int base=1e9+7;
int f

;
int a
;

void add(int &a, int b)
{
    if((a+=b)>=base) a-=base; 
}

void work()
{
    int i,j,n,p;
    
    scanf("%d%d",&n,&p);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        a[i]%=p;
    }
    
    memset(f,0,sizeof(f));    
    f[0][0]=1;
    for(i=1;i<=n;i++)
        for(j=0;j<p;j++)
        {
            add(f[i][j],f[(i-1)][j]);
            add(f[i][j],f[(i-1)][(j-a[i]+p)%p]);
        }
    printf("%d\n",f
[0]);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
        work();
    return 0;
}