牛客，小红不想做完全背包

https://ac.nowcoder.com/acm/contest/78904/D

题意：n个物品，求物品的价值和为p的倍数的方案中，最少几件物品。

思路：完全背包，将背包容量从p扩大到p的20倍，可以直接ac。
看了题解，发现用bfs做的，感觉逻辑也还行。
想尝试用物品预处理出所有可能的余数，并记录到达该余数的最小数量值，然后当背包处理。但是不知道为什么不能ac，代码先贴上。

void solve(){
    int n, p;
    cin >> n >> p;

    const int inf = 0x3f3f3f3f;

    vector<int> weights(p + 1, inf);
    for (int i = 1; i <= n; ++i){
        long long value;
        cin >> value;
        value %= p;
        weights[value] = 1;
        int k = 2;
        while (k < p){
            int cur = value * k % p;
            //cout << cur << " " << k << " " << weights[cur] << endl;
            weights[cur] = min(weights[cur], k);
            k ++;
            //cout << cur << " " << k << " " << weights[cur] << endl;
        }
    }
    //for (int i = 0; i <= p; ++i){cout << weights[i] << " \n"[i == p];}
    vector<int> dp = weights;
    dp[0] = 0;
    dp[p] = weights[0];
    for (int i = 1; i < p; ++i){
        if (weights[i] == inf){
            continue;
        }
        for (int j = i; j <= p; ++j){
            dp[j] = min(dp[j], dp[j - i] + min(weights[i], dp[i]));
        }
    }

    cout << dp[p] << endl;
   // for (int i = 0; i <= p; ++i){cout << weights[i] << " \n"[i == p];}
}