Huge number (C language)

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 .