题目：
HPLC法中色谱柱的理论塔板数用于评价（）。
A.分析时间
B.保留时间
C.柱效
D.柱压
E.流速
答案：
参考答案：C
HPLC法中色谱柱的理论塔板数用于评价（）。 A.分析时间 B.保留时间 C.柱效
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题型：单项选择题 A1/A2型题
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题型：问答题
[说明]
假定用一个整型数组表示一个长整数，数组的每个元素存储长整数的一位数字，则实际的长整数m表示为：
m=a[k]×10k-2+a[k-1]×10k-3+…+a[3]×10+a[2]
其中a[1]保存该长整数的位数，a[0]保存该长整数的符号：0表示正数、1表示负数。
运算时先决定符号，再进行绝对值运算。对于绝对值相减情况，总是绝对值较大的减去绝对值较小的，以避免出现不够减情况。注意，不考虑溢出情况，即数组足够大。
[函数]
int cmp(int *LA, int *LB);
/*比较长整数LA与LB的绝对值大小*/
/*若LA绝对值较大返回正值，LA较小返回负值，相等则返回0*/
int ADD (int *LA, int *LB, int *LC)
/*计算长整数LA与LB的和，结果存储于LC中*/
/*注意：正数与负数的和相当于正数与负数绝对值的差*/
/*数据有误返回0，正常返回1*/
if(LA == NULL || LB == NULL || LC == NULL)return 0;
int *pA, *pB, i, N, carry, flag;
flag = LA[0] + LB[0];
switch(flag) /*根据参与运算的两个数的符号进行不同的操作*/
case 0:
case 2:
Lc[0] = LA[0];/*LA与LB同号，结果符号与LA(LB)相同*/
pA = LA;
pB = LB;
(1) ;
break;
case 1: /*LA与LB异号*/
/*比较两者的绝对值大小，结果符号与较大者相同*/
flag = (2) ;
if(flag > 0) /*LA较大*/
LC[0] = LA[0];
pA = LA;
pB = LB;
else if(flag < 0)(/*LB较大*/
LC[0] = LB[0];
pA = LB;
pB = LA;
else/*LA与LB相等*/
LC[0] = 0;
LC[1] = 0;
return 1;
flag = -1;
break;
default:
return 0;
break;
/*switch*/
/*绝对值相加减*/
/*注意对于减法pA指向较大数，pB指向较小数，不可能出现不够减情况*/
(3) ;
N = LA[1] > LB[1] LA[1] : LB[1];
for(i = 0; i < N; i++)
if(i >= pA[1])/*LA计算完毕*/
carry += flag * pB[i+2];
else if(i >= pB[1])/*LB计算完毕*/
carry += pA[i+2];
else
carry += pA[i+2] + flag * pB[i+2];
LC[i+2] = carry % 10;
carry /= 10;
if( (4) )/*需要借位，针对减法*/
LC[i+2] += 10;
carry--;
/*for*/
if( (5) )/*最高进位，针对加法*/
LC[i+2] = carry;
i++;
if(LC[i+1] == 0) i--; /*若最高位为零，针对减法*/
LC[1] = i;
return 1;
;/*ADD*/
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