void combi(bool val[MAX], int curr_combi[9], int limit_use, int limit_range, int curr_arg, int curr_uses, int sum){
if(limit_use == curr_uses || limit_range+1 == curr_arg){
val[sum] = true;
return;
}
combi(val,curr_combi,limit_use,limit_range,curr_arg,curr_uses+1,sum+curr_combi[curr_arg]);
combi(val,curr_combi,limit_use,limit_range,curr_arg+1,curr_uses+1,sum+curr_combi[curr_arg]);
combi(val,curr_combi,limit_use,limit_range,curr_arg+1,curr_uses,sum);
}
void combi(char *s, char *s1)
{
char t[100] = {0};
if (strlen(s) == 0)
{
printf("%s\n", s1);
return;
}
strcpy(t, s1);
combi(s+1, t);
int n = strlen(t);
t[n] = s[0]; t[n+1] = '\0';
combi(s+1, t);
}
//--------------------------------------------------------------
void ofApp::draw(){
ofDrawBitmapString("nCm % x == 0:light",10,50);
ofDrawBitmapString("nCm % x != 0:dark",10,40);
ofDrawBitmapString("push 1 key:x=1",10,30);
ofDrawBitmapString("push any key:x++",10,20);
ofDrawBitmapString("x:"+ofToString(x),10,10);
double center = STEP * SIZE / 2;
for(int i = 0 ;i <= STEP; i++){
for (int j = 0; j <= i; j++) {
int c = combi(i, j);
//枠
ofSetColor(0,0,0);
ofRect((i/2.0-j)*SIZE + center, i*SIZE, SIZE, SIZE);
//中身
ofSetColor(64+200*(c%x==0),64+200*(c%x==0),64+200*(c%x==0));
ofRect((i/2.0-j)*SIZE + center+1.5, i*SIZE+1.5, SIZE-3, SIZE-3);
//数字
ofSetColor(0, 0, 0);
string text = ofToString(c);
//string text = string((4-(int)log10(c))/2,' ') + ofToString(c);
//std::cout<<(4-(int)log10(c))<<":"<<text<<std::endl;
ofDrawBitmapString(text,(i/2.0-j)*SIZE+center+SIZE/9, i*SIZE+SIZE*3/5);
}
}
}
// s 表示要排列的資料
// n 表示此陣列的大小
// m 表示要選出的數量
// pos 表示要排列pos ~ n-1之間選出m-got個資料
// got 表示已經選好的個數
void combi(char *s, int n, int m, int pos, int got) {
int i;
// 請問終止條件
if (m == got) {
// 印出來看對不對
for (i = 0; i < m; i++)
printf("%c", s[i]);
printf("\n");
return;
}
//如何化簡？
//由pos道n-1間可以任選出i位置的元素
for (i = i; i <= n-1; i++) {
//如何選定？選定到哪個位置？
//應該選到 got 這個位置上！
//若採用 swap(got,i) 的方式選擇, 那程式應該這樣寫
tmp = s[i];
s[i] = s[got];
s[got] = tmp;
//接下來如何recursion?
//由 i+1 開始選, 因為組合和位置無關, 前面選過的組合不能再選
/* Example
A B C D E c(5, 2)
A -> AB, AC, AD, AE
B -> BC ,BD, BE
C -> CD, CE
E -> DE
*/
combi(s, n, m, i+1 , got+1);
//recursion 完要記得還原陣列
tmp = s[i];
s[i] = s[got];
s[got] = tmp;
}
}
int paint()
{
int i, n, r;
for(n = 0; n <= N; n++)
{
for(r = 0; r <= n; r++)
{
if(0 == r) //打印空格
{
for(i = 0; i <= (N-n); i++)
{
printf(" ");
}
}
else
{
printf(" ");
} //打印空格结束
printf("%3d", combi(n,r));
}
printf("\n");
}
return 0;
}
int main() {
array();
tuple();
unordered_map();
unordered_set();
forward_list();
combi();
}
main()
{
char s[100], s1[100];
printf("Enter the string : ");
scanf("%s", s);
strcpy(s1, s);
permut(s, 0);
combi(s1, "");
}
void combinator_main(){
int n, r;
for(n = 0; n <= 5; n++){
for(r = 0; r <= n; r++){
printf("%d C %d=%ld ", n, r, combi(n, r));
}
printf("\n");
}
}
int main(void)
{
int n,r;
for (n = 0; n <= 5; n++) {
for (r = 0; r <= n; r++)
printf("%dC%d=%ld " ,n,r,combi(n,r));
printf("\n");
}
}
int main()
{
int t, n, r;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &n, &r);
long long int ans = combi(n, r);
printf("%lld\n", ans);
}
return 0;
}
int make_limit(int curr_combi[9], int curr_pos){
bool val[MAX];
for(int i=0;i<MAX;i++) val[i]=false;
for(int i=0;i<=curr_combi[curr_pos];i++) val[i]=true;
for(int i=1;i<=num_stamp;i++)
combi(val,curr_combi,i,curr_pos,0,0,0);
for(int i=1;i<MAX;i++) if(val[i]==false) return i;
}
void main(void)
{
int n, r, t;
for (n = 0; n <= N; n++){
for (t = 0; t < (N - n) * 3; t++){
printf(" ");
}
for (r = 0; r <= n; r++){
printf("%3ld ", combi(n, r));
}
printf("\n");
}
}
lld combi(lld n, lld k,lld m)
{
if(n<k)
return 0;
if(n-k<k)
return combi(n,n-k,m);
lld i,p=1,t=1;
for(i=n-k+1;i<=n;i++)
p=(p*i)%m;
for(i=1;i<=k;i++)
t=(t*i)%m;
return (p*inverse_modulo(t,m))%m;
}
int main(void)
{
int n,r,t;
for (n=0;n<=N;n++)
{
for (r=0;r<=n;r++)
{
int i; //排版设定开始
if (r==0)
{
for (i=0;i<=(N-n);i++)
printf(" ");
}
else
printf(" "); //排版设定结束
printf("%3d",combi(n,r));
}
printf("\n");
}
return 0;
}
void paint()
{
int n, r, t;
for(n = 0; n <= N; n++)
{
for(r = 0; r <= n; r++)
{
int i;/* 排版设定开始*/
if(r == 0)
{
for(i = 0; i <= (N-n); i++)
printf(" ");
}
else
{
printf(" ");
} /* 排版设定结束*/
printf("%3d", combi(n, r));
}
printf("\n");
}
}
long combi(int n, int r) {
if(r == 0 || r == n)
return 1L;
else
return combi(n - 1, r) + combi(n - 1, r - 1);
}
int combi(int a,int b){//aCb
if(a == b || b == 0) return 1;
return combi(a-1,b) + combi(a-1,b-1);
}
bool btSubsimplexConvexCast::calcTimeOfImpact(
const btTransform& fromA,
const btTransform& toA,
const btTransform& fromB,
const btTransform& toB,
CastResult& result)
{
btMinkowskiSumShape combi(m_convexA,m_convexB);
btMinkowskiSumShape* convex = &combi;
btTransform rayFromLocalA;
btTransform rayToLocalA;
rayFromLocalA = fromA.inverse()* fromB;
rayToLocalA = toA.inverse()* toB;
m_simplexSolver->reset();
convex->setTransformB(btTransform(rayFromLocalA.getBasis()));
//btScalar radius = btScalar(0.01);
btScalar lambda = btScalar(0.);
//todo: need to verify this:
//because of minkowski difference, we need the inverse direction
btVector3 s = -rayFromLocalA.getOrigin();
btVector3 r = -(rayToLocalA.getOrigin()-rayFromLocalA.getOrigin());
btVector3 x = s;
btVector3 v;
btVector3 arbitraryPoint = convex->localGetSupportingVertex(r);
v = x - arbitraryPoint;
int maxIter = MAX_ITERATIONS;
btVector3 n;
n.setValue(btScalar(0.),btScalar(0.),btScalar(0.));
bool hasResult = false;
btVector3 c;
btScalar lastLambda = lambda;
btScalar dist2 = v.length2();
#ifdef BT_USE_DOUBLE_PRECISION
btScalar epsilon = btScalar(0.0001);
#else
btScalar epsilon = btScalar(0.0001);
#endif //BT_USE_DOUBLE_PRECISION
btVector3 w,p;
btScalar VdotR;
while ( (dist2 > epsilon) && maxIter--)
{
p = convex->localGetSupportingVertex( v);
w = x - p;
btScalar VdotW = v.dot(w);
if ( VdotW > btScalar(0.))
{
VdotR = v.dot(r);
if (VdotR >= -(SIMD_EPSILON*SIMD_EPSILON))
return false;
else
{
lambda = lambda - VdotW / VdotR;
x = s + lambda * r;
m_simplexSolver->reset();
//check next line
w = x-p;
lastLambda = lambda;
n = v;
hasResult = true;
}
}
m_simplexSolver->addVertex( w, x , p);
if (m_simplexSolver->closest(v))
{
dist2 = v.length2();
hasResult = true;
//printf("V=%f , %f, %f\n",v[0],v[1],v[2]);
//printf("DIST2=%f\n",dist2);
//printf("numverts = %i\n",m_simplexSolver->numVertices());
} else
{
dist2 = btScalar(0.);
}
}
//int numiter = MAX_ITERATIONS - maxIter;
// printf("number of iterations: %d", numiter);
result.m_fraction = lambda;
result.m_normal = n;
return true;
}