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Huge number (C language)
2022-07-04 10:27:00 【Lol only plays Timo】
What is a huge number ？ seeing the name of a thing one thinks of its function , Huge numbers are especially large numbers , As big as C Language is no longer recognizable , For example, thousands of trillions of addition, subtraction, multiplication and division ,C Linguistic int, And long int Obviously, it can't be expressed , Because they are all four byte complements , The range of integer numbers that can be represented is -2,147,483,648 ~ 2,147,483,647, namely -2^32 ~ 2^32-1.
So how to solve this problem ？
First , according to int We can use Ten thousand system , Similar to decimal , It's just that the decimal is 4 One by one , The decimal system is a group of digits . Speaking of this, some people may be confused , Why is it ten thousand decimal , Not 100000 , Or thousands , Choose a decimal . Because if we are doing multiplication , Everyone is at most 9999 The multiplication of the two will not exceed int Representation range of , It's not true int The representation range of causes too much waste .
Now let's solve the problem of digital storage ：
We can save the numbers into the file in advance , Transfer directly from the file , This is convenient for our testing , And it is convenient for us to modify , After the data is transferred from the file, the next step is to convert these a lot of strings into hexadecimal and store them in a int In an array of types .
Here's a tip , It's the opposite , In other words, the low order of a huge number is stored in the front of the array , The higher the bit, the lower the stored subscript , The advantage of doing this is that it is very consistent with our computing habits when performing operations, starting from the low order , And it is also subscripted from the array 0 Start with the number of . Then let's take a look at the code ：
void storeInfor(HUGE_NUM *NUM, char *temp) {
int i = 0;
int j = 0;
int k = 0;
int flag = 1;
if (temp[0] == '-') {
NUM->sign = 1;
for (i = 0; i < NUM->count-1; i++) {
temp[i] = temp[i+1];
}
NUM->count--;
} else {
NUM->sign = 0;
}
for (j = NUM->count-1, i = 0; j >= NUM->count%4 + 3; i++) {
NUM->value[i] = ((int)temp[j]-48) + ((int)temp[j-1] - 48) * 10 + ((int)temp[j-2] - 48)* 100 + ((int)temp[j-3] - 48)* 1000;
j = j-4;
}
for (k = NUM->count%4-1,flag = 1; k >= 0; k--) {
if (k == NUM->count%4-1) {
NUM->value[i] = ((int)temp[k]-48) * flag;
} else {
NUM->value[i] += ((int)temp[k]-48) * flag;
}
flag = flag * 10;
}
printInfor(NUM);
}
It doesn't matter if you can't understand some of the code here , Finally, I will attach all the complete code , For your reference .
At this point, we have saved all the figures , Let's first look at the basic operations - Addition and subtraction .
Of course, the addition operation is very simple , Here you can first try how to realize the addition operation of decimal .
Compared with addition , Subtraction is much harder to achieve , Because of carry, we can easily add one to him , But excuse me , If there is continuous borrowing, it will be a big test for us , What's going on ？
Here I would like to introduce a way studied by my teacher Zhu Hong - Wechat complement , According to the underlying principle knowledge , Inferential , We all know that computer subtraction is realized by complement , My teacher's method is the same . Next, let's take a look at this hand of God ：
Now these main knowledge and technologies have been told to you , Let's take a look at my code ：
#include <stdio.h>
#include <malloc.h>
typedef struct HUGE_NUM {
int sign;
int *value;
int count;
}HUGE_NUM;
void storeInfor(HUGE_NUM *NUM, char *temp);
void addNum(HUGE_NUM *NUM1, HUGE_NUM *NUM2, HUGE_NUM *NUM3, char *temp1, char *temp2);
void printInfor(HUGE_NUM *NUM);
void printInfor(HUGE_NUM *NUM) {
int i;
int count = (NUM->count+3) / 4 - 1;
if (NUM->sign == 1) {
printf("-:");
} else {
printf("+:");
}
if(NUM->value[(NUM->count+3) / 4 - 1] == 0) {
count = (NUM->count+3) / 4 - 2;
}
for (i = count; i >= 0; i--) {
if (i == count) {
printf("%d ",NUM->value[i]);
} else {
printf("%04d ",NUM->value[i]);
}
}
printf("\n");
}
void addNum(HUGE_NUM *NUM1, HUGE_NUM *NUM2, HUGE_NUM *NUM3, char *temp1, char *temp2) {
int sign;
int *caculate1;
int *caculate2;
int count;
int i;
if (NUM1->count >= NUM2->count) {
count = (NUM1->count + 3) / 4 + 1;
caculate1 = (int *) calloc(sizeof(int),count);
caculate2 = (int *) calloc(sizeof(int),count);
} else {
count = (NUM2->count + 3) / 4 + 1;
caculate1 = (int *) calloc(sizeof(int),count);
caculate2 = (int *) calloc(sizeof(int),count);
}
for (i = 0; i < (NUM1->count+3)/4; i++) {
caculate1[i] = NUM1->value[i];
}
for (i = 0; i < (NUM2->count+3)/4; i++) {
caculate2[i] = NUM2->value[i];
}
if (NUM1->sign == NUM2->sign) {
if(NUM1->sign == 0) {
NUM3->sign = 0;
} else {
NUM3->sign = 1;
}
for (i = 0; i < count; i++) {
if (i == 0) {
NUM3->value[i] = (caculate1[i]+caculate2[i])%10000;
} else {
NUM3->value[i] = (caculate1[i]+caculate2[i])%10000 + (caculate1[i-1]+caculate2[i-1])/10000;
if(NUM3->value[i-1] == 10000) {
NUM3->value[i]++;
NUM3->value[i-1] = 0;
}
}
}
if(NUM3->value[count-1] >= 10000) {
NUM3->value[count-1] -= 10000;
NUM3->value[count]++;
}
} else {
if (NUM1->sign == 1) {
for (i = 0; i < count; i++) {
caculate1[i] = 9999 - caculate1[i];
}
}
if (NUM2->sign == 1) {
for (i = 0; i < count; i++) {
caculate2[i] = 9999 - caculate2[i];
}
}
for (i = 0; i < count; i++) {
if (i == 0) {
NUM3->value[i] = (caculate1[i]+caculate2[i])%10000;
} else {
NUM3->value[i] = (caculate1[i]+caculate2[i])%10000 + (caculate1[i-1]+caculate2[i-1])/10000;
if(NUM3->value[i-1] == 10000) {
NUM3->value[i]++;
NUM3->value[i-1] = 0;
}
}
}
if(NUM3->value[count-1] >= 10000) {
NUM3->value[count-1] -= 10000;
NUM3->value[count]++;
}
if (NUM3->value[count] == 1) {
NUM3->value[0] += 1;
NUM3->sign = 0;
}
if (NUM3->value[count] == 0) {
NUM3->sign = 1;
for (i = 0; i < count; i++) {
NUM3->value[i] = 9999 - NUM3->value[i];
}
}
}
}
void storeInfor(HUGE_NUM *NUM, char *temp) {
int i = 0;
int j = 0;
int k = 0;
int flag = 1;
if (temp[0] == '-') {
NUM->sign = 1;
for (i = 0; i < NUM->count-1; i++) {
temp[i] = temp[i+1];
}
NUM->count--;
} else {
NUM->sign = 0;
}
for (j = NUM->count-1, i = 0; j >= NUM->count%4 + 3; i++) {
NUM->value[i] = ((int)temp[j]-48) + ((int)temp[j-1] - 48) * 10 + ((int)temp[j-2] - 48)* 100 + ((int)temp[j-3] - 48)* 1000;
j = j-4;
}
for (k = NUM->count%4-1,flag = 1; k >= 0; k--) {
if (k == NUM->count%4-1) {
NUM->value[i] = ((int)temp[k]-48) * flag;
} else {
NUM->value[i] += ((int)temp[k]-48) * flag;
}
flag = flag * 10;
}
printInfor(NUM);
}
int main () {
FILE *fp1;
FILE *fp2;
char temp1[10000];
char temp2[10000];
int i = 0;
int j = 0;
int k = 0;
int flag = 1;
HUGE_NUM NUM1;
HUGE_NUM NUM2;
HUGE_NUM NUM3;
fp1 = fopen("testdata1.txt","r");
fp2 = fopen("testdata2.txt","r");
fseek(fp1,0,SEEK_END);
NUM1.count = ftell(fp1);
fseek(fp2,0,SEEK_END);
NUM2.count = ftell(fp2);
if (NUM1.count >= NUM2.count) {
NUM3.count = NUM1.count + 1;
} else {
NUM3.count = NUM2.count + 1;
}
NUM1.value = (int *) calloc(sizeof(int), ((NUM1.count+3) / 4));
NUM2.value = (int *) calloc(sizeof(int), ((NUM2.count+3) / 4));
NUM3.value = (int *) calloc(sizeof(int), ((NUM3.count+3) / 4));
fseek(fp1,0,SEEK_SET);
fseek(fp2,0,SEEK_SET);
fgets(temp1,1000,fp1);
fgets(temp2,1000,fp2);
storeInfor(&NUM1,temp1);
storeInfor(&NUM2,temp2);
addNum(&NUM1, &NUM2, &NUM3, temp1, temp2);
printInfor(&NUM3);
fclose(fp1);
fclose(fp2);
return 0;
}
It's not very good , You can point out any problems , Welcome to leave a message below , The multiplication program will be updated later .
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