大数操作，用顺序表实现最大一百位数的加减法实现

#include <stdio.h>
#include <stdlib.h>

// 最大一百位数的加减法实现
const int MAXSIZE = 100;

typedef struct {

int length;						//长度
int numbers[MAXSIZE];			//各个位上的数字，从低位数字到高位数字
int sign;						//符号：0代表正数，1代表负数，规定0的符号为正

}SqList;

//头插，即新元素插入到numbers[0]
void insert(SqList *l, int x)
{
SqList *p = l;

if(l->length<MAXSIZE){

int i;
//所有后移
for(i=l->length-1;i>=0;i--){
l->numbers[i+1]=l->numbers[i];
}

l->numbers[0] = x;
l->length++;
}

}

//初始化： s：符号，0正1负；a：包含从高位到低位各个位数的数组；n：位数，也即数组a的长度
SqList *init(int s, int *a, int n)
{
SqList *L;
L=(SqList*)malloc(sizeof(SqList));
if(!L)
{
printf("Init Error!");
exit(1);
}
L->sign=s;
L->length=0;

int i;
for(i=0;i<n;i++){
insert(L,a[i]);
}
return L;
}

void log(SqList *l){

if(l!=NULL){

printf("-------------\n");
printf("length: %d\n",l->length);
printf("sign: %d\n", l->sign);
printf("numbers: ");

int i=0;

for(i=l->length-1;i>=0;i--){
printf("%d ", l->numbers[i]);
}
printf("\n");
}

}

// 简单的正数相加
SqList *add(SqList *l1, SqList *l2){

SqList *l = init(0,0,0);

int c = 0;		//进位

int i;
int t;		//temp value for sequence add.

int len = (l1->length>l2->length)?l2->length:l1->length;

//按照加法逻辑，从低位到高位开始相加
for(i=0;i<len;i++){
t =	l1->numbers[i] + l2->numbers[i] + c;

l->numbers[i] = t%10;
l->length++;

if(t>=10){
c = t/10;
}else{
c = 0;
}
}

//按照加法逻辑，以下处理较大数字的高位
len = l1->length-l2->length;
len = (len<0)?-len:len;
len = len+i;

for(i;i<len;i++){

printf("true\n");

int t;

if(l1->length>l2->length){
t = l1->numbers[i] + c;
}else{
t = l2->numbers[i] + c;
}

l->numbers[i] = t%10;
l->length++;

if(t>=10){
c = t/10;
}else{
c = 0;
}
}

return l;
}

// 简单的正数相减，不论顺序，永远返回较大数减去较小数的值
SqList *sub(SqList *l1, SqList *l2){

SqList *l = init(0,0,0);
int c = 0;		// 借位

int i;
int t;

int len = (l1->length>l2->length)?l2->length:l1->length;

//按照减法逻辑，从低位到高位开始相减
for(i=0;i<len;i++){

if(l1->length>l2->length){
if((l1->numbers[i]-c)<l2->numbers[i]){
t = l1->numbers[i]+10 - l2->numbers[i] - c;
c = 1;
}else{
t =	l1->numbers[i] - l2->numbers[i] - c;
c = 0;
}

}else{

if((l2->numbers[i]-c)<l1->numbers[i]){
t = l2->numbers[i]+10 - l1->numbers[i] - c;
c = 1;
}else{
t =	l2->numbers[i] - l1->numbers[i] - c;
c = 0;
}

}

l->numbers[i] = t%10;
l->length++;

}

//处理较大数的高位
len = l1->length-l2->length;
len = (len<0)?-len:len;
len = len+i;

for(i;i<len;i++){

int t;

if(l1->length>l2->length){
t = l1->numbers[i] - c;
}else{
t = l2->numbers[i] - c;
}

//如果和借位相减为0，得数应该减少一位
if(t==0){
l->length--;
}

l->numbers[i] = t%10;
l->length++;

if(t<0){
c = 1;
}else{
c = 0;
}
}

return l;
}

//绝对值比较，用于真正的加减运算判断
int cmpabs(SqList* l1, SqList *l2){
if(l1->length!=l2->length){
return (l1->length>l2->length)?(1):(0);
}else{
int i;
for(i=l1->length-1;i>=0;i--){
if(l1->numbers[i]>l2->numbers[i]){
return 1;
}else if(l1->numbers[i]<l2->numbers[i]){
return 0;
}
}
return 2;
}

}

//加法，判断两数符号做相应处理
SqList* ADD(SqList* l1, SqList* l2){
if(l1->sign == 0 && l2->sign == 0){
return add(l1,l2);
}else if(l1->sign == 1 && l2->sign == 1){
SqList* l = add(l1,l2);
l->sign = 1;
return l;
}else{

SqList* l = sub(l1,l2);
l->sign = !cmpabs(l1,l2);
return l;
}

}

//减法，判断两数符号做相应处理，较加法而言还要考虑操作数顺序
SqList* SUB(SqList* l1, SqList* l2){

int t = cmpabs(l1,l2);
if(t == 2){
int a[1] = {0};
SqList* l = init(0,a,1);
return l;
}

if(l1->sign == 1 && l2->sign == 1){
if(t == 0){
return sub(l1,l2);
}else if(t == 1){
SqList* l = sub(l1,l2);
l->sign = 1;
return l;
}

}else if(l1->sign == 0 && l2->sign == 0){

if(t == 1){
return sub(l1,l2);
}else if(t == 0){
SqList* l = sub(l1,l2);
l->sign = 1;
return l;
}
}else if(l1->sign == 1 && l2->sign == 0){
SqList* l = add(l1,l2);
l->sign = 1;
return l;

}else if(l1->sign == 0 && l2->sign == 1){

return add(l1,l2);
}

}

int main()
{

printf("Application starts...\n");

int a1[3] = {1,2,2};
SqList *l1 = init(1, a1, 3);
log(l1);

int a2[2] = {2,4};
SqList *l2 = init(0, a2 ,2);
log(l2);

SqList *l3 = init(1, a2 ,2);

SqList *l = SUB(l2,l1);
log(l);

return 0;

}