Ejemplo n.º 1
0
int main(){
	int n;
	//因此当模m有原根时,它有φ(φ(m))个原根。
	while( scanf("%d", &n) != EOF ){
		printf("%d\n", eular(eular(n)));
	}
	return 0;
}
Ejemplo n.º 2
0
int main() {
    int m, k;
    int i;
    int total;
    while(~scanf("%d%d", &m, &k)) {
        if(m == 1)
            printf("%d\n", k);
        else {
            long long res = eular(m);
            long long cur = k / res;
            if(k % res == 0)
                cur--;
            k = k - cur * res;
            cur = cur * m + 1;
            long long i;
            long long temp;
            for(i = cur; k != 0; i++) {
                if(gcd(i, m) == 1) {
                    temp = i;
                    --k;
                }
            }
            printf("%lld\n", temp);
        }
    }
}
	// 根据摄像机方向改变billboard的纹理
	bool BillboardSetObject::objectRendering(const MovableObject *mo, const Camera *camera)
	{
		const Quaternion &camQ = m_billboardSet->getCamQ();
		Vector3 eular(MathEx::QuaternionToEular(camQ));
		

		return true;
	}
Ejemplo n.º 4
0
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        printf("%d\n", eular(n - 1));
    }
    return 0;
}
Ejemplo n.º 5
0
void eular(int u) {
    used[u] = 1;
    int i;
    while (head[u] != -1) {
        i = head[u];
        head[u] = Next[head[u]];
        if (vis[i])continue;
        vis[i ^ 1] = 1;
        eular(To[i]);
    }
}
Ejemplo n.º 6
0
int main()
{
    long long n;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&n);
    printf("%lld\n",eular(n+1));
    }
    return 0;
}
Ejemplo n.º 7
0
    void solve(std::istream &in, std::ostream &out) {
        int t;
        in >> t;
        while (t--) {
            int n, m;
            in >> n >> m;
            init(n);
            rep(i, n)
            deg[i] = 0;
            int u, v;
            rep(i, m)
            {
                in >> u >> v;
                u--, v--;
                deg[u]++;
                deg[v]++;
                addedge(u, v);
                addedge(v, u);
            }
            gao = -1;
            rep(i, n)
            {
                if (deg[i] & 1) {
                    if (gao != -1) {
                        addedge(i, gao);
                        addedge(gao, i);
                        gao = -1;
                    } else gao = i;
                }
            }
            rep(i, n)
            used[i] = 0;
            rep(i, n)
            {
                if (!used[i]) {
                    eular(i);
                }
            }
            m <<= 1;
            for (int i = 0; i < m; i += 2) {
                if (!vis[i])out << 1 << endl;
                else out << 0 << endl;
            }

        }
int print_eular(double x_0, double y_0, double x_n, double step)
{
	double res = eular(x_0,y_0,x_n,step);
	printf("The result of eular method is : %lf \n\r",res);
	return 0;
}