本文整理汇总了C++中product函数的典型用法代码示例。如果您正苦于以下问题：C++ product函数的具体用法？C++ product怎么用？C++ product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product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: interact
// only c1 and c2 can be modified
static void interact(const struct cell *c1, const struct cell *c2) {
if (is_leaf(c1) && is_leaf(c2)) {
direct(c1, c2);
} else if (is_leaf(c1)) {
map(interact, product(c1, get_subcells(c2)));
} else if (is_leaf(c2)) {
map(interact, product(get_subcells(c1), c2));
} else {
map(interact, product(get_subcells(c1), get_subcells(c2)));
}
}
开发者ID:naoyam，项目名称:tapas，代码行数:12，代码来源:direct.c
示例2: normalize_vec4
void normalize_vec4(vec4 *v)
{
scalar n;
/* normalize v */
product(v, v, &n);
if ((n != (scalar) 0.0) && (n != (scalar)1.0)) {
n = (scalar)1.0 / sqrtf(n);
product(&n, v, v);
}
}
开发者ID:rpmsousa，项目名称:humanoid，代码行数:11，代码来源:linear.c
示例3: power
Poly power(Poly P, LL n, const Poly &coef)
{
Poly ret(P.size());
ret[0] = 1;
while (n) {
if (n & 1) ret = product(ret, P, coef);
P = product(P, P, coef);
n >>= 1;
}
return ret;
}
开发者ID:neroneroni，项目名称:ACM-ICPC，代码行数:11，代码来源:linear_rec_seq.cpp
示例4: multinomialSamplingProb
long double multinomialSamplingProb(const vector<long double>& probs, const vector<int>& obs) {
vector<long double> factorials;
vector<long double> probsPowObs;
factorials.resize(obs.size());
transform(obs.begin(), obs.end(), factorials.begin(), factorial);
vector<long double>::const_iterator p = probs.begin();
vector<int>::const_iterator o = obs.begin();
for (; p != probs.end() && o != obs.end(); ++p, ++o) {
probsPowObs.push_back(pow(*p, *o));
}
return factorial(sum(obs)) / product(factorials) * product(probsPowObs);
}
开发者ID:YuankunZhu，项目名称:freebayes，代码行数:12，代码来源:Multinomial.cpp
示例5: v3
v3 v3::transform_4(mat44 t_mat){
colvec4 operand = {x, y, z, c};
colvec4 product = t_mat * operand;
// if (product(3) == -7 && c==1){
// std::cout << "operand --" << std::endl;
// operand.print();
// std::cout << "product --" << std::endl;
// product.print();
// std::cout << "t_mat --" <<std::endl;
// t_mat.print();
// }
return v3(product(0), product(1), product(2), product(3));
}
开发者ID:suBDavis，项目名称:raster，代码行数:13，代码来源:v3.cpp
示例6: main
int main(int argc, char *argv[]) {
FILE* output;
double a[nmax], b[nmax], c[nmax];
double sa, sb, sc, max;
int na, nb, nc;
int minval, maxval;
setlocale(LC_ALL, "rus");
// Проверки на правильность введенных аргументов
if (argc < 6) {
fprintf(stderr, "Недостаточно параметров!\n");
return -1;
}
if (!sscanf(argv[1], "%d", &minval)){
fprintf(stderr, "1-ый и 2-ой аргумент должны представлять минимальную и максимальную границу диапозона, соответственно.\n");
return -2;
}
if (!sscanf(argv[2], "%d", &maxval)){
fprintf(stderr, "1-ый и 2-ой аргумент должны представлять минимальную и максимальную границу диапозона, соответственно.\n");
return -3;
}
if (minval >= maxval){
fprintf(stderr, "Пустой диапозон.\n");
return -4;
}
if ((output = fopen(s_output, "w")) == NULL) {
fprintf(stderr, "Невозможно создать файл '%s'\n", s_output);
return -5;
}
// Вывод диапазона
fprintf(output, "Диапазон: %d - %d\n", minval, maxval);
// Ввод массивов из файлов
if (!ArrayInput(&na, a, argv[3], &fclose))
return -5;
if (!ArrayInput(&nb, b, argv[4], &fclose))
return -6;
if (!ArrayInput(&nc, c, argv[5], &fclose))
return -7;
// Умножение элементов, выходящих за границы
sa = product(a, na, minval, maxval, &check);
sb = product(b, nb, minval, maxval, &check);
sc = product(c, nc, minval, maxval, &check);
// Поиск максимального значения
max = sa;
if (sb > max) max = sb;
if (sc > max) max = sc;
// Вывод результатов
if (sa == max) fprintf(output, "Массив А имеет максимальное произведение элементов: %9.3lf\n", max);
if (sb == max) fprintf(output, "Массив B имеет максимальное произведение элементов: %9.3lf\n", max);
if (sc == max) fprintf(output, "Массив C имеет максимальное произведение элементов: %9.3lf\n", max);
}
开发者ID:vdsirotkin，项目名称:10_programs，代码行数:50，代码来源:lab6.cpp
示例7: return
bool Interval::intersect_soft (Interval tmp)
{
if ((product(tmp.s).sign() == product(tmp.e).sign() && (product(tmp.s)).sign() != 0) ||
(tmp.product(s).sign() == (tmp.product(e)).sign() && (tmp.product(s)).sign() != 0))
return false;
if (product(tmp.s).sign() == product(tmp.e).sign() && (product(tmp.s)).sign() == 0)
{
Fraction max1;
Fraction max2;
Fraction min1;
Fraction min2;
if(s.getX() != e.getX())
{
max1 = (s.getX() <= e.getX() ? e.getX() : s.getX());
max2 = (s.getX() <= e.getX() ? e.getX() : s.getX());
min1 = (s.getX() <= e.getX() ? s.getX() : e.getX());
min2 = (s.getX() <= e.getX() ? e.getX() : s.getX());
}
else
{
max1 = (s.getY() <= e.getY() ? e.getY() : s.getY());
max2 = (s.getY() <= e.getY() ? e.getY() : s.getY());
min1 = (s.getY() <= e.getY() ? s.getY() : e.getY());
min2 = (s.getY() <= e.getY() ? e.getY() : s.getY());
}
return (max1 >= min2 && max2 >= min1);
}
if(product(tmp.s).sign() != product(tmp.e).sign() && tmp.product(s).sign() != (tmp.product(e)).sign())
return true;
return false;
}
开发者ID:nitroman1991，项目名称:Division，代码行数:31，代码来源:subsidiary.cpp
示例8: product
char *sum(void){
char *var1, *var2;
var1 = product();
while(match(PLUS)){
advance();
var2 = product();
//printf("%s = %s + %s\n", var1,var1,var2);
//printf("ADD %s, %s\n", var2,var1);
fprintf(f,"ADD %s, %s\n", var2,var1);
freename(var2);
}
return var1;
}
开发者ID:sanketh95，项目名称:compilers，代码行数:14，代码来源:code_gen_new.c
示例9: InvMixCol
static u32 InvMixCol(u32 x) { /* matrix Multiplication */
u32 y,m;
u8 b[4];
m=pack(InCo);
b[3]=product(m,x);
m=ROTL24(m);
b[2]=product(m,x);
m=ROTL24(m);
b[1]=product(m,x);
m=ROTL24(m);
b[0]=product(m,x);
y=pack(b);
return y;
}
开发者ID:DatMatt，项目名称:dop-mii，代码行数:15，代码来源:rijndael.c
示例10: InvMixCol
/*列混合的逆变换*/
static WORD InvMixCol(WORD x)
{ /* matrix Multiplication */
WORD y,m;
BYTE b[4];
m=pack(InCo);
b[3]=product(m,x);
m=ROTL24(m);
b[2]=product(m,x);
m=ROTL24(m);
b[1]=product(m,x);
m=ROTL24(m);
b[0]=product(m,x);
y=pack(b);
return y;
}
开发者ID:RUIXUEZHAOFENGNIAN，项目名称:mycode，代码行数:16，代码来源:AES加解密算法1.cpp
示例11: main
int main()
{
FILE *in = fopen("ride.in", "r");
FILE *out = fopen("ride.out", "w");
char comet[LEN], group[LEN];
fscanf(in, "%s\n%s", comet, group);
fprintf(out, "%s\n", (product(comet)%47 == product(group)%47) ? "GO" : "STAY");
fclose(in);
fclose(out);
return 0;
}
开发者ID:brunocasteleiro，项目名称:competitive-programming，代码行数:15，代码来源:ride.c
示例12: norm
void CLyapWolfMethod::orthonormalize()
{
if (mNumExp < 1) return;
//TODO generalize
C_FLOAT64 *dbl, *dblEnd;
dbl = mVariables.array() + mSystemSize;
dblEnd = dbl + mSystemSize;
mNorms[0] = norm(dbl, dblEnd);
scalarmult(dbl, dblEnd, 1 / mNorms[0]);
size_t i, j;
for (i = 1; i < mNumExp; ++i)
{
dbl += mSystemSize;
dblEnd = dbl + mSystemSize;
//orthogonalisation
for (j = 0; j < i; ++j)
{
add(dbl, dblEnd,
-product(dbl, dblEnd, mVariables.array() + (j + 1)*mSystemSize),
mVariables.array() + (j + 1)*mSystemSize);
}
//normalisation
mNorms[i] = norm(dbl, dblEnd);
scalarmult(dbl, dblEnd, 1 / mNorms[i]);
}
}
开发者ID:jonasfoe，项目名称:COPASI，代码行数:32，代码来源:CLyapWolfMethod.cpp
示例13: load_nc_array
template<class T> static bool load_nc_array(const NcFile& ncf, const string& name, vector<T>& dest, bool required = true, int offset = 0, int count = -1)
{
NcVar *v = load_nc_variable(ncf, name.c_str(), required);
if (v)
{
vector<long> offsets = list_of(offset).repeat(v->num_dims()-1, 0);
v->set_cur(&offsets.front());
vector<long> counts (v->num_dims());
long* shape = v->edges();
transform(shape, shape + v->num_dims(), offsets.begin(), counts.begin(), minus<long>());
delete shape;
if (count > 0)
{
counts[0] = count;
}
dest.resize(product(counts));
bool success = v->get(&dest.front(), &counts.front());
if (!success)
{
dest.resize(0);
check(!required, string("NetcdfDataset::load_nc_array<") + typeid(T).name() + "> " + name + '\n' + "failed with offset " + str(offsets) + ", counts " + str(counts));
}
return success;
}
return false;
}
开发者ID:tpajenka，项目名称:RNNLIB，代码行数:26，代码来源:NetcdfDataset.hpp
示例14: productExcludeItself
/**
* @param A: Given an integers array A
* @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]
*/
vector<long long> productExcludeItself(vector<int> &nums) {
if (nums.size() < 2) {
return vector<long long>();
}
vector<long long> left_product(nums.size());
vector<long long> right_product(nums.size());
vector<long long> product(nums.size());
left_product[0] = 1;
for (int i = 1; i < nums.size(); ++i) {
left_product[i] = left_product[i - 1] * nums[i - 1];
}
right_product[nums.size() - 1] = 1;
for (int j = nums.size() - 2; j >= 0; --j) {
right_product[j] = right_product[j + 1] * nums[j + 1];
}
for (int k = 0; k < nums.size(); ++k) {
product[k] = left_product[k] * right_product[k];
}
return product;
}
开发者ID:codingwhite，项目名称:LintCode，代码行数:29，代码来源:product-of-array-exclude-itself.cpp
示例15: product
UMatrix<T>& UMatrix<T> :: operator*(const T mult)
{
// if(m_dimension != source.m_dimension) throw Exception(5);
UMatrix<T> product(m_dimension);
for(int i=0; i<m_dimension; i++)
{
for(int j=0; j<m_dimension-i; j++)
{
product(i,j) = m_ptr_to_data[i][j] * mult;
}
}
copy(product);
return *this;
}
开发者ID:cwang218，项目名称:CPP，代码行数:16，代码来源:umatrix.hpp
示例16: init_fb
bool
init_fb(const struct grid_info grid)
{
bool ret = true;
if (grid.stages & GL_COMPUTE_SHADER_BIT) {
const struct image_info img = image_info_for_grid(grid);
const unsigned n = product(grid.size) *
image_num_components(grid.format);
uint32_t *pixels = malloc(n * sizeof(*pixels));
ret = init_pixels(img, pixels, 0.5, 0.5, 0.5, 0.5) &&
upload_image(img, max_image_units(), pixels);
free(pixels);
} else {
ret = generate_fb(grid, 0);
glClearColor(0.5, 0.5, 0.5, 0.5);
glClear(GL_COLOR_BUFFER_BIT);
glClearDepth(0.5);
glClear(GL_DEPTH_BUFFER_BIT);
}
return ret;
}
开发者ID:chemecse，项目名称:piglit，代码行数:27，代码来源:common.c
示例17: multiply
vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
int row1 = A.size();
int col1 = A[0].size();
int row2 = B.size();
int col2 = B[0].size();
vector<vector<int>> product(row1, vector<int>(col2, 0));
for(int i = 0; i < row1; i++) {
bool zeroRow = true;
for(int p = 0; p < col1; ++p) {
if(A[i][p] != 0) {
zeroRow = false;
break;
}
}
if(zeroRow) continue;
for(int j = 0; j < col2; ++j) {
bool zeroCol = true;
for(int q = 0; q < row2; ++q) {
if(B[q][j] != 0) {
zeroCol = false;
break;
}
}
if(zeroCol) continue;
for(int k = 0; k < col1; ++k) {
if(A[i][k] == 0 or B[k][j] == 0) continue;
product[i][j] += (A[i][k] * B[k][j]);
}
}
}
return product;
}
开发者ID:Kaidul，项目名称:LeetCode_problems_solution，代码行数:33，代码来源:Sparse_Matrix_Multiplication.cpp
示例18: main
int main(int argc, char const *argv[])
{
int pro;
pro=product(5,1,2,3,4,5);
printf("%d\n",pro );
return 0;
}
开发者ID:hejiangda，项目名称:C_Program_Test，代码行数:7，代码来源:ex14.2.c
示例19: main
int main()
{
float a, b, r;
char op;
do {
printf("Extra line of code --- \n");
printf("Number operator Number: ");
scanf(" %f %c %f", &a, &op, &b);
switch (op)
{
case '+' : r = add(a,b);
break;
case '*' : r = product(a,b);
break;
case '-' : r = subtract(a,b);
break;
case '/' : r = divide(a,b);
break;
case '%' : r = remainder(a,b);
break;
case 'q' : break;
default : op='?';
}
if (op=='?')
printf("Unknown operator\n");
else if (op=='q')
printf("Bye\n");
else
printf("%f %c %f = %f\n", a, op, b, r);
}
while (op != 'q');
return 0;
}
开发者ID:KishanUnnadkat，项目名称:maths_function，代码行数:35，代码来源:maths_functions.c
示例20: main
int main(int argc, char *argv[]) {
int fds[2] = {0};
int th = 0;
int ret = pipe(fds);
if (ret < 0) {
printf("pipe failed\n");
return -1;
}
th = fork();
if (th < 0) {
printf("fork failed\n");
return -1;
} else if (th == 0) {
product(fds);
} else {
consumer(fds);
}
wait(NULL);
return 0;
}
开发者ID:Jeff-Yu，项目名称:fju，代码行数:28，代码来源:pipe.c
注：本文中的product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。 |
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