SQL Server - 根据周数获取一周内的第一个日期？(SQL Server - Get first date in a week, given the week number?)

 首先，你需要一个函数whitch计算表中一个区间内的所有频率： 
 
int countinterval( double *tab, int tabsize, double min, double max )
{
    int count = 0;
    for ( int i = 0; i < tabsize; i ++ )
    {
        if ( tab[i] >= min && tab[i] < max ) // test if frequence is in interval [min,max[
        // if ( tab[i] >= min && tab[i] <= max ) // test if frequence is in interval [min,max]
            count ++;
    }
    return count;
}
 
 接下来，您需要一个函数whitch汇总每个表的结果。 
 
int counttabsinterval( double *tab[], int tabsize, int tabcount, double min, double max )
{
    int count = 0;
    for ( int i = 0; i < tabcount; i ++ )
        count += countinterval( tab[i], tabsize, min, max ); // sums results
    return count;
}
 
 在您想知道频率数的每个间隔调用该函数。 
 
double *tab[1024];
int tabsize = ???; // <- init with size of one table
int count1 = counttabsinterval( tab, tabsize, 1024, -1.0, -0.9 );
int count2 = counttabsinterval( tab, tabsize, 1024, -0.9, -0.8 );
 
 这是一个计算表格tab上的定量频率分布的函数，其中包含表格的长度，并将结果存储在具有长度distcount的数组dist 。 
 
void QuantitativeFrequencyDistribution( double *tab, int tabsize, double *dist, int distcount )
{
    int count = 0;
    double intervallAmount = 2.0 / distcount;
    for ( int i = 0; i < distcount; i ++ )
    {                            
        double min = -1.0 + i*intervallAmount;
        double max = min + intervallAmount;
        dist[i] = countinterval( tab, tabsize, min, max );
    }
}

int sizeOfTab = ?;// <- init with size of table
double *tab; // <- your table
int distcount = 20; // for example
double dist[distcount];
QuantitativeFrequencyDistribution( tab, sizeOfTab, dist, distcount );
 
 如果您有不同长度的不同表，并且您想知道所有表的定量频率分布，则必须为每个表调用QuantitativeFrequencyDistribution并计算结果的总和。 

First you need a function which counts all frequences in an interval in on table:
 
int countinterval( double *tab, int tabsize, double min, double max )
{
    int count = 0;
    for ( int i = 0; i < tabsize; i ++ )
    {
        if ( tab[i] >= min && tab[i] < max ) // test if frequency is in interval [min,max[
        // if ( tab[i] >= min && tab[i] <= max ) // test if frequency is in interval [min,max]
            count ++;
    }
    return count;
}
 
Next you need a function which sums the results of each table.
 
int counttabsinterval( double *tab[], int tabsize, int tabcount, double min, double max )
{
    int count = 0;
    for ( int i = 0; i < tabcount; i ++ )
        count += countinterval( tab[i], tabsize, min, max ); // sums results
    return count;
}
 
Call the function for each interval where you want to know number of frequencies.
 
double *tab[1024];
int tabsize = ???; // <- init with size of one table
int count1 = counttabsinterval( tab, tabsize, 1024, -1.0, -0.9 );
int count2 = counttabsinterval( tab, tabsize, 1024, -0.9, -0.8 );
 
Here is a function which calculates the Quantitative Frequency Distribution for on table tab with lenght of table tabsize ans stores the result in an array distwith length distcount.
 
void QuantitativeFrequencyDistribution( double *tab, int tabsize, double *dist, int distcount )
{
    int count = 0;
    double intervallAmount = 2.0 / distcount;
    for ( int i = 0; i < distcount; i ++ )
    {                            
        double min = -1.0 + i*intervallAmount;
        double max = min + intervallAmount;
        dist[i] = countinterval( tab, tabsize, min, max );
    }
}

int sizeOfTab = ?;// <- init with size of table
double *tab; // <- your table
int distcount = 20; // for example
double dist[distcount];
QuantitativeFrequencyDistribution( tab, sizeOfTab, dist, distcount );
 
If you have different tables with different lengths and you want to know the Quantitative Frequency Distribution over all tables, you have to call QuantitativeFrequencyDistribution for each single table and calculate sum of the results.