int main()
{
	long n,k,l,a,b;
	scanf("%ld%ld%ld",&k,&l,&n);
	a=mx(calc(k,l,n),calc(k,0,n-1));
	b=mn(calc2(k,l,n),calc2(0,l,n-1)+3);
	printf("%ld %ld\n",a,b);
	return 0;
}
Пример #2
0
double RPNStackCalc(stack stk)
{
	int i;
	node item;
	stack calc;
	memset(&calc,0,sizeof(calc));
	calc.size=0;
	for(i=0;i<stk.size;i++)
	{
		item=stk.item[i];
		if(item.is_num)
		{
			push(&calc,item);
		}else{
			node a,b;
			nodecpy(&b,pop(&calc));
			nodecpy(&a,pop(&calc));
			item.data.num=calc2(a.data.num,b.data.num,item.data.op);
			item.is_num=true;
			push(&calc,item);
		}
	}
	if(calc.size!=1||isEmpty(calc)) 
		return -1;
	else 
		return (calc.item[0].data.num);
}
void calc3()
{
	int l;
	for(l=1;l<m3;l++)
	{
		calc2(0,0,0,0,l);
		//calc2(0,1,0,0,l);
		//calc2(0,0,1,0,l);
		//calc2(0,0,0,1,l);
	}
}
Пример #4
0
void solve() {
	int M;
	scanf("%d", &M);
	int ans = 0;
	ans += calc1(M);
	if ( ans <= 1 )
		ans += calc2(M);
	if ( ans <= 1 )
		ans += calc3(M);
	printf("%s\n", ans == 1 ? "YES" : "NO");
}
Пример #5
0
void calc4ee(struct exp6 left, int r1, int r2)
{
    struct exp6 ret = calc2(r1, r2);
    int i, c;
    enum opr op;

    for (i = 0; i < ret.c; i++)
        for (c = 0; c < left.c; c++)
            for (op = 0; op < 4; op++)
                calc4ee_(left.d[c], op, ret.d[i]);
}
Пример #6
0
void calc4(int a, int b, int c, int d)
{
    struct exp6 left = calc2(a,b);
    calc4ee(left, c, d);
    calc4enn(left, c, d);
}