5 line Python, 100%

Works for arbitrary number of coins with arbitrary values. Standard dp, use memoization or @cache python builtin.

def solve(n, coins, v=[1, 2, 5, 10]):
    @cache
    def dp(s, p):
        if s==0 or p==0:
            return 1 if s<=coins[p] else 0
        return sum(dp(s-i*v[p], p-1) for i in range(min(coins[p], s//v[p])+1))
    return dp(n, len(coins))

Python small solution.

def f(n,x,i,z,dp):
    if n==0:return 1
    if i>3 :return 0
    if (n,i) in dp:return dp[n,i]
    ans=0
    for j in range(x[i]+1):
        if n-j*z[i]<0:break
        ans+=f(n-j*z[i],x,i+1,z,dp)
    dp[n,i]=ans
    return ans
def solve(n, x):
    z=(1,2,5,10)
    return f(n,x,0,z,{})

SHORTEST CODE TO THIS QUESTION

int min(int a, int b) {int c=a; if(a>b) c=b; return c; }

int main() { int t,n,a,b,c,d,i,j,k,l; scanf("%d",&t);

while(t--)
{int x=0;
    scanf("%d %d %d %d %d",&n,&a,&b,&c,&d);

    for(i=min(d,n/10);i>=0;i--)
    {
        for(j=min((n-10*i)/5,c);j>=0;j--)
        {
            for(k=min(b,(n-10*i-5*j)/2);k>=0;k--)
            {
                for(l=min(a,(n-10*i-5*j-2*k));l>=0;l--)
                {
                    if(10*i+5*j+2*k+l==n)
                    {
                        x++;
                        break;
                    }
                }
            }
        }
    }
    printf("%d\n",x);
}
return 0;

}

Solution for C++

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int modd=1e9;
int T,n,a,b,c,d;
ll ans;
int main() {
	ios::sync_with_stdio(false);
	cin>>T;
	while(T--) {
		cin>>n,ans=0;
		cin>>a>>b>>c>>d;
		for(int i=0; i<=min(d,n/10); i++)
			for(int j=0; j<=min(c,n/5); j++)
				for(int k=0; k<=min(b,n/2); k++)
					if(n-10*i-5*j-2*k<=a&&n-10*i-5*j-2*k>=0) ans++;
		cout<<ans<<endl;
	}
	return 0;
}

The editorial optimization 3 is wrong and bad way . The optimization 4 is good approach . Suppose u have 0 1 coins , 3 2 coins ,2 5 coins , 1 10 coins. Then U can't make N if(13) But as rule of optimization of 3

(2*i+5*j+10*k>=(N-A)) count++;

many times 2*i+5*j+10*k this part will be greater than (13-0) . So what rubbish solution that is !!!

include

using namespace std;

int main() { int T,N,A,B,C,D; cin>>T; while(T--) { cin>>N; cin>>A>>B>>C>>D; long long ways[N+1]={0}; for(int i=0;i<=A;i++) { for(int j=0;j<=B&&(i+2*j)<=N;j++) { ways[(i+2*j)]++; } } long long ans=0; for(int i=0;i<=C&&(i*5)<=N;i++) { for(int j=0;j<=D&&((i*5)+(10*j))<=N;j++) { ans+=ways[N-((i*5)+(10*j))]; } } cout<