让我们假设你所有的 NI 是 1 。

让方法[J] =的获取和J的方式数。

您可以计算这个像这样（这是你在做什么，但我不知道为什么你叫你的变量素数）。

 办法[0] = 1
对于i = 1到m做
    对于j = myLim DOWNTO X [I]做
        方法[J] + =方法[J  -  X [I]];
 

这意味着你只使用值的每个硬币曦键。您可以添加另一个循环至少一次，至多 NI 次但使用它：

 办法[0] = 1
对于i = 1到m做
    对于时间= 1到n [I]做//使用习一次，然后两次，然后三，...，然后妮
        对于j = myLim DOWNTO倍* X [I]做
            方法[J] + =方法[J  - 次* X [I]];
 

您仍然可以申请你的模数，并计算你的极限，我离开了那些对简便性。

I have written a program for generating subset sum which might be used in this problem which states:

Suppose, you have 3 $1-coins, 2 $2-coins, 3 $5-coins, 1 $10-coin, there are 4 ways to obtain $10 from those coins. If there are n1 $X1 coins, n2 $X2 coins.... nm $Xm coins, how many ways can we obtain $X from these limited number of coins?

If we create a set of { X1, X1..... X1, X2, X2.......... X2, ..., ..., ............, Xm, Xm... Xm}, and then run Subset summing on it, surely we can get a result for $X. But I could not find a way to use the sets {n1, n2, n3.... nm} , {X1, X2, X3.... Xm}. A friend told me that it is a variation of knapsack problem, but I am not sure, how.

This is a partial code of what I have written :

ways[0]=1, mylim=0;
for(i=0;i<count;i++){
    if(mylim+coins[i]<=LIMIT) mylim+=coins[i];
    else mylim=LIMIT;

    for(j=mylim; j>=coins[i];j--){
        ways[j]=(ways[j]+ways[j-coins[i]])%MOD;
    }
}

It would be great for me if you are kind enough to explain a bit elaborately.

EDIT: This question is more suitable for stackexchange for computer science, but since it is an old question of mine, I'm rather editing it here.

This problem can be solved with Inclusion Exclusion principle, and it comes handy when we have coin values fixed but number of each coin varying with each query.

Suppose, ways[v] is ways of making $v with $x1, $x2, .. $xm, each being used as many times as needed. Now, if we are using only n1 numbers of $x1, we have to subtract the configurations using at least (n1 + 1) numbers of $x1 ( which is actually ways[v - (n1 + 1)x1] ). Moreover, if we are using only n2 numbers of $x2, we have to subtract ways[v - (n2 + 1)x2] as well, and etc.

Now, we have twice subtracted the configurations where at least (n1 + 1) $x1 and (n2 + 1) $x2 are used, hence we need to add ways[v -(n1 + 1)x1 - (n2 + 1)x2] and etc.

In particular, if,

N = set of configurations where all coins are used as many as possible,

Ai = set of configurations where at least ni + 1 numbers of $xi is used, for 1 <= i <= m, then

the result we are seeking = |N| - |A1| - |A2| .. - |Am| + |A1 and A2| + |A1 and A3| + ... - |A1 and A2 and A3| .....

The code which computes number of configurations with unlimited coins is actually simpler:

ways[0]=1;
for( int i = 0 ; i < count ; i++){
    for( int j = coins[i] ; j < ways.size() ; j++ ){
        ways[j] += ways[j-coins[i]];
    }
}

Let's assume all your ni are 1.

Let ways[j] = number of ways of obtaining sum j.

You can compute this like so (this is what you're doing, but I don't know why you named your variable primes).

ways[0] = 1
for i = 1 to m do
    for j = myLim downto X[i] do
        ways[j] += ways[j - X[i]];

This means you only use each coin of value Xi once. You can add another loop to use it at least once and at most ni times however:

ways[0] = 1
for i = 1 to m do
    for times = 1 to n[i] do // use Xi one time, then two times, then three, ..., then ni
        for j = myLim downto times*X[i] do
            ways[j] += ways[j - times*X[i]];

You can still apply your modulo and compute your limit, I left those out for simplicity.