• 设为首页
  • 点击收藏
  • 手机版
    手机扫一扫访问
    迪恩网络手机版
  • 关注官方公众号
    微信扫一扫关注
    公众号

C++ comb函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了C++中comb函数的典型用法代码示例。如果您正苦于以下问题:C++ comb函数的具体用法?C++ comb怎么用?C++ comb使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。



在下文中一共展示了comb函数的20个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的C++代码示例。

示例1: cal_delta

int cal_delta(int h,int alpha,int beta)    //calculate the delta of inter_polynomial, c is inter_poly[i] and t is tk
{
	int delta=0,delta_temp=0,i,j,flag;
	int a,b,x,y;

	x=inter_point[0];			//xi=pi
	y=inter_point[1];			//yi=ri

	for(j=0;j<poly_Ysize; j++)
	{
		for(i=0;i<poly_Xsize;i++)
		{
			if( (inter_poly[h][j][i] != 0)  && (j>=beta) && (i>=alpha) )
			{
				a=i;
				b=j;
				flag= ((int)comb(a,alpha)) * ((int)comb(b,beta));
				// if flag is even number , delta_temp is equal to 0
				// if flag is odd number, delta_temp can be calculated to the delta
				if( (flag%2) !=0 )
				{
					delta_temp=mul(inter_poly[h][j][i],power(x,(a-alpha)));
					delta_temp=mul(delta_temp,power(y,(b-beta)));
					delta= add(delta,delta_temp);
				}
			}
		}	
	}

	return delta;
}
开发者ID:codywsy,项目名称:Simulation-by-C,代码行数:31,代码来源:[cheating]RS(63,15)_m=1.cpp


示例2: comb

void comb(int s, int n, int m)
{
    int i;

    if (s >= n || flag==1)
        return ;

    if (top == m)
    {

        if (flag==queue[m-2] && m>1)
            return ;

        for (i = 0; i < m; i++)
        {
            printf("%d", queue[i]);
            ss = ss + e0[queue[i]] ;
        }
        printf("\t");
        if( (1-sigma/temperature * (ss + (NPARTICLE - m) * e0[0])) > 0)
        {
        // for (i = 0; i < m; i++)
        // {
        //     printf("%d", queue[i]);
        // }
        // printf("\t");
            partition[q-1] = partition[q-1] + pow((1-sigma/temperature * (ss + (NPARTICLE - m) * e0[0])),1/sigma);
            ppartition[q-1] = ppartition[q-1] + pow((1-sigma/temperature * (ss + (NPARTICLE - m) * e0[0])),1/sigma+1);
        // printf("(1-sigma/temperature*(ss+(NPARTICLE-m)*e0[0])^(1/sigma)=");
        printf("%2.2e",pow((1-sigma/temperature * (ss + (NPARTICLE - m) * e0[0])),1/sigma) );
        printf("\t");
        printf("%2.2e\n",(1-sigma/temperature * (ss + (NPARTICLE - m) * e0[0])) );
            flag=queue[m-2];
        flag=0;
        }
        else
        {
            // goto jump;
            partition[q-1] = partition[q-1] + 0.0;
            ppartition[q-1] = ppartition[q-1] + 0.0;
        printf("\t");
        printf("\t");
        printf("%2.2e\n",(1-sigma/temperature * (ss + (NPARTICLE - m) * e0[0])) );
            flag=1;
        }
        sum=sum+1;
        ss = 0.0;
        return ;
    }

    queue[top++] = array[s];

    // if (flag==queue[m-2])
    //     return ;

    comb(s, n, m);
    top--;
    comb(s+1, n, m);

}
开发者ID:JackXu1993,项目名称:Finite_System_Statistical_Mechanics,代码行数:60,代码来源:cleast.c


示例3: comb

int comb(int n,int r)
{
	if(r==0||r==n)
	    return 1;
	else
	return comb(n-1,r-1)+comb(n-1,r);
}
开发者ID:sparsh,项目名称:concepts-and-algorithms,代码行数:7,代码来源:comination.c


示例4: main

int main()
{
	int monsters;
	int skills;
	printf("How many monsters do you have?: " );
	scanf("%i", &monsters);
	printf("How many skillups do you want?: ");
	scanf("%i", &skills);

	long long combinations = comb(monsters, skills);
	
//	printf("Done with comb");
	long double answer = 0;
	for(int i = skills; i <= monsters; i++)
	{
		answer += noSkill(monsters - i) * yesSkill(i) * comb(monsters, i);	
		printf("%Lf answer\n", answer);
	}
	answer *= 100;
		
//	double answer = noSkill*yesSkill*combinations*100;
	//printf("%f noSkill\n", noSkill);
	//printf("%f yesSkill\n", yesSkill);
	
	printf("You have a %Lf%% chance to get at least  %i skillups out of %i monsters\n", answer, skills, monsters);

	return -1;
}
开发者ID:DerrickChanCS,项目名称:COEN12,代码行数:28,代码来源:skillup.c


示例5: comb

void comb(int s, int n, int m)
{
    int i;

    if (s >= n)
        return ;

    if (top == m)
    {
        for (i = 0; i < m; i++)
        {
            printf("%d", queue[i]);
            // ss = ss + queue[i] ;
        }
            // printf("%d", ss);
        printf("\t");
        sum=sum+1;
        ss = 0.0;
        return ;
    }

    queue[top++] = array[s];
    comb(s, n, m);
    top--;
    comb(s+1, n, m);
}
开发者ID:JackXu1993,项目名称:Finite_System_Statistical_Mechanics,代码行数:26,代码来源:combination_recursive.c


示例6: main

//===========================
int main(void) {
	setvbuf(stdout, NULL, _IONBF, 0);
	FILE *outPutFile; 
	int n;
	int r;
	char buffer[BUFSIZ+1];
	 outPutFile = fopen("comb.txt", "w");
	puts("Combinatorials"); /* prints !!!Hello World!!! */
	n = 0;
	r = 0;
	do{
		printf("Enter value one:");
		fprintf(outPutFile, "Enter value one:");
		n = atoi(gets(buffer));
		fprintf(outPutFile,"%d", n);
		//printf("\n");
		printf("Enter value two:");
		fprintf(outPutFile, "Enter value two:");
		r = atoi(gets(buffer));
		fprintf(outPutFile,"%d", r);
		//printf("\n");
		if(check(n, r)){
			printf("comb %d\n", comb(n, r));
			fprintf(outPutFile,"comb %d\n", comb(n, r));
		}
	}while((r > 0)&&( n > 0));
	
	
	return EXIT_SUCCESS;
}
开发者ID:sbcc-computer-science,项目名称:cclass,代码行数:31,代码来源:TAndersonCombinatorials.c


示例7: comb

int comb(int a, int n)
{
    if(a == 0) return 1;
    if(a < 0) return 0;
    if(n == 0) return 0;
    
    return comb(a, n - 1) + comb(a - coins[NUM - n][1], n);
}
开发者ID:Yegorov,项目名称:myCppReference,代码行数:8,代码来源:coins_exchange.c


示例8: comb

long long comb(int i, int j)
{
	if(i<0||j<0) return 1;
	if(Comb[i][j])
		return Comb[i][j];
	if(!i||!j||i==j)
		return Comb[i][j]=1;
	return Comb[i][j]=comb(i-1,j)+comb(i-1,j-1);
}
开发者ID:dementrock,项目名称:acm,代码行数:9,代码来源:7098994_AC_0MS_224K.cpp


示例9: comb

int comb (int a, int b) {
  if (dp[a][b] != 0)
    return dp[a][b];
  if (a < b)
    return -1;
  if (a == b || b == 0) {
    dp[a][b] = 1;
  } else {
    dp[a][b] = comb(a-1, b-1) + comb(a-1, b);
  }
  return dp[a][b];
}
开发者ID:kw-udon,项目名称:ucc,代码行数:12,代码来源:combination.c


示例10: comb

unsigned long long int comb(int n,int k)
{
        if(arr[n][k])
        {
                return arr[n][k];
        }
        if(k==1  || n-k==1)
                return n;
        if(n==k)
                return 1;
        return (arr[n][k]=comb(n-1,k-1)+comb(n-1,k));
}
开发者ID:abhisheknith,项目名称:Code,代码行数:12,代码来源:spoon.c


示例11: comb

//setter
void Curve::setCurve(vector<Point> points) {
	table.clear();
	int n = points.size();
 	int x, y, u;
  	for (float i=0;i<1;i+=0.0001) {
    	x = 0; y = 0;
    	for (u=0;u<n;u++) {
      		x += comb(n-1,u)*pow(i,u)*pow(1-i,n-1-u)*points[u].getAbsis();
      		y += comb(n-1,u)*pow(i,u)*pow(1-i,n-1-u)*points[u].getOrdinat();
    	}
    	table.push_back(Point(x,y));
  	}
}
开发者ID:fauzannaufan,项目名称:Grafika06,代码行数:14,代码来源:Curve.cpp


示例12: comb

void comb(int pool, int need, marker chosen, int at)
{
	if (pool < need + at) return; /* not enough bits left */
 
	if (!need) {
		/* got all we needed; print the thing.  if other actions are
		 * desired, we could have passed in a callback function. */
		for (at = 0; at < pool; at++)
			if (chosen & (one << at)) combs = combs + 1;
		return;
	}
	/* if we choose the current item, "or" (|) the bit to mark it so. */
	comb(pool, need - 1, chosen | (one << at), at + 1);
	comb(pool, need, chosen, at + 1);  /* or don't choose it, go to next */
}
开发者ID:chris-wood,项目名称:llvm-pgo,代码行数:15,代码来源:combinations.c


示例13: main

void main() {
	IplImage* img;
	CvCapture* cap=cvCaptureFromCAM(0);
	cvNamedWindow("Line Counter", 1);
	CvFont* font1=new CvFont;
	CvFont* font2=new CvFont;
	cvInitFont(font1, CV_FONT_HERSHEY_SIMPLEX, 0.5f, 1.0f, 0, 3, 8);
	cvInitFont(font2, CV_FONT_HERSHEY_SIMPLEX, 0.5f, 1.0f, 0, 2, 8);
	int val=0, axx=0, bxx=0;
	char text[8];
	for (;;) {
		img = cvQueryFrame(cap);
		if (!img) break;
		IplImage* gray1=cvCreateImage(cvSize(img->width, img->height), 8, 1);
		IplImage* edge1=cvCreateImage(cvSize(img->width, 16), 8, 1);
		cvCvtColor(img, gray1, 7);
		extract(gray1, edge1);
		dy(edge1, edge1);
		cvThreshold(edge1, edge1, 10, 255, CV_THRESH_BINARY_INV);
		val=count(edge1);
		if (val==0&&axx==0) { axx=1; }
		if (val==2&&axx==1) { axx=0; bxx++; }
		sprintf(text, "%i", bxx);
		comb(gray1, edge1);
		cvPutText(gray1, text, cvPoint(10, 160), font1, cvScalarAll(255));
		cvPutText(gray1, text, cvPoint(10, 160), font2, cvScalarAll(0));
		cvShowImage("Line Counter", gray1);
		if (cvWaitKey(5) > 0) break;
		cvReleaseImage(&gray1);
		cvReleaseImage(&edge1);
	}
}
开发者ID:ArdWar,项目名称:Kuliah,代码行数:32,代码来源:LineCounter.cpp


示例14: main

int main(void) {
	int Nj = 4;
	
	unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
	std::mt19937_64 gen(seed);
	std::uniform_real_distribution<float> dis(0, 1);
	
	std::vector<float> v;
	for(int i = 0; i < Nj; ++i) v.push_back(dis(gen));
	
	std::cout << "individual probabilities:" << std::endl;
	for(auto val: v) std::cout << std::fixed << val << "\t";
	std::cout << std::endl << std::endl;
	
	float sum = 0.0;
	for(int Ntag = 0; Ntag <= Nj; ++Ntag) {
		std::cout << "combined probabilities:" << std::endl;
		float combi = comb(v, Nj, Ntag);
		std::cout << std::endl << "the sum of combined probability:\t";
		std::cout << combi << std::endl;
		sum += combi;
	}
	
	std::cout << std::endl << "overall sum of probabilities:\t" << sum << std::endl;
	
	return EXIT_SUCCESS;
}
开发者ID:HEP-KBFI,项目名称:Summies2014-KE,代码行数:27,代码来源:combinations.cpp


示例15: gen

 int gen(int m, int n, pair<int, int>& p, unordered_map<string, vector<string>>& source, unordered_map<string, string>& refer) {
     string pos = to_string(p.first) + "#" + to_string(p.second);
     refer[pos] = pos;
     source[pos] = vector<string>(1, pos);
     
     vector<string> List;
     if (p.first > 0) {
         string tempPos = to_string(p.first-1) + "#" + to_string(p.second);
         List.push_back(tempPos);
     }
     if (p.first < m-1) {
         string tempPos = to_string(p.first+1) + "#" + to_string(p.second);
         List.push_back(tempPos);
     }
     if (p.second > 0) {
         string tempPos = to_string(p.first) + "#" + to_string(p.second-1);
         List.push_back(tempPos);
     }
     if (p.second < n-1) {
         string tempPos = to_string(p.first) + "#" + to_string(p.second+1);
         List.push_back(tempPos);
     }
     for (string l : List) comb(source, refer, l, pos);
     
     return source.size();
 }
开发者ID:Dauphin-NWU,项目名称:Leetcode-Practice,代码行数:26,代码来源:305_number_of_islands_2.cpp


示例16: combinationSum2

 vector<vector<int> > combinationSum2(vector<int> &num, int target) {
     vector<vector<int> > ret;
     vector<int> temp;
     sort(num.begin() , num.end());
     comb(ret , temp , num , 0 , target);
     return ret;
 }
开发者ID:mabodx,项目名称:leetcode,代码行数:7,代码来源:Combination+Sum+II.cpp


示例17: main

int main()
{
    printf("k=%d", comb(100, 5));
    //int n = 0;
    //printf("%d", coins[NUM - (n)][1]);
    return 0;
}
开发者ID:Yegorov,项目名称:myCppReference,代码行数:7,代码来源:coins_exchange.c


示例18: n

nBinEqvCod::nBinEqvCod(int n, int w, int q) : n(n), w(w), q(q)
{
    omp_set_dynamic(1);
    omp_set_num_threads(8);
    mpz_ui_pow_ui(qw.get_mpz_t(), (q - 1), w);
    //qw = qPow((q - 1), w);
	//qw = Bigint(q-1).pow(w);

	M = mpz_class(qw) * comb(n, w); // factorial(n) / (factorial(w) * factorial(n - w));

    //code = new nBinEqvVec[M];
    code = nullptr;

	qDebug() << "Creating non_binary equivalent codes";
	qDebug() << "n:" << n << "\tw: " << w << "\tq:" << q << "\tM: " << M.get_str().c_str();

	//#pragma omp parallel for
	//for (int i = 0; i < M; ++i)
	//{
	//	calc_eVec(i, &code[i]);
	//	mapNBEV.insert(code[i].Ca, &code[i]);
	//}
	//--------------------------------------------------
	//QTime midnight(0, 0, 0);
	//qsrand(midnight.secsTo(QTime::currentTime()));
	//code = new nBinEqvVec[3];
	//calc_eVec(qrand() % M, code[0]);
	//calc_eVec(qrand() % M, code[1]);
	//calc_eVec(qrand() % M, code[2]);
}
开发者ID:Quill88,项目名称:FieldsGalua_and_NBinEqvCod_Qt,代码行数:30,代码来源:nBinEqvCod.cpp


示例19: judge

int judge(int m, int l)                 // m=particles number on the excited state, l=data point
{
    int i,j;                             // loop index
    int lmax;                            // maximum energy level


    for (j=1; j<MAXIMUM_ENERGY; j++)
    {
        if((1 - ( (PARTICLE_NUMBER - m) * e0[0] + (m-1) * e0[1] + e0[j])/(temperature[l]*N3))< 0)
        {
            lmax = j - 1;
            printf("%d\n", lmax);
            printf("%4.2e\n", (1 - ( (PARTICLE_NUMBER - m) * e0[0] + (m-1) * e0[1] + e0[j-1])/(temperature[l]*N3)));
            break;
        }
        else
        {
            lmax = j;
        }
    }

    if (lmax==0)
    {
        return 0;
    }
    else
    {
        comb(lmax, m, l);
        return 1;
    }
}
开发者ID:JackXu1993,项目名称:Finite_System_Statistical_Mechanics,代码行数:31,代码来源:delta(no).c


示例20: main

int main()
{
	long int a,b,total;
	scanf("%ld %ld",&a,&b);
	total=comb(a,b,1,1);
	printf("%ld\n",total);
	return 0;
}
开发者ID:vijaykeswani,项目名称:Online-Judge,代码行数:8,代码来源:comb.c



注:本文中的comb函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


鲜花

握手

雷人

路过

鸡蛋
该文章已有0人参与评论

请发表评论

全部评论

专题导读
上一篇:
C++ combine函数代码示例发布时间:2022-05-30
下一篇:
C++ com函数代码示例发布时间:2022-05-30
热门推荐
阅读排行榜

扫描微信二维码

查看手机版网站

随时了解更新最新资讯

139-2527-9053

在线客服(服务时间 9:00~18:00)

在线QQ客服
地址:深圳市南山区西丽大学城创智工业园
电邮:jeky_zhao#qq.com
移动电话:139-2527-9053

Powered by 互联科技 X3.4© 2001-2213 极客世界.|Sitemap