加减乘的深度应用

C语言求高精度加减乘

用数组来模拟加减乘除的运算，加法是关键

加法：

int main()

{

//创建两个数组，用来存放两个加数

char arr1[1000] = {"0"};

char arr2[1000] = { "0" };

int arr3[1001] = {0};

scanf("%s %s",arr1,arr2);

int ch1[1000] = {0};

int ch2[1000]={0};

int i = 0;

int len1 =strlen(arr1);

int len2 =strlen(arr2);

int len = len1 >len2 ? len1 : len2;

for (i = len1 - 1; i>= 0; i--)

{

ch1[i]= arr1[len1-i-1]-48;//使数组逆序排列，进行加法，因为加法从地位到高位相加

}

for (i = len2 - 1; i>= 0; i--)

{

ch2[i]= arr2[len2-i-1]-48;//同理

}

int arrlen = len+1;

for (i = 0; i <arrlen; i++)

{

//核心算法

arr3[i]+= ch1[i] + ch2[i];

arr3[i+ 1] = arr3[i] / 10;

arr3[i]%= 10;

}

while ((arr3[arrlen] == 0&& arrlen > 0))

arrlen--;

//printf("%d ", arrlen);

for (i = arrlen; i>= 0; i--)

printf("%d", arr3[i]);

return 0;

}

减法

//Swap函数用来交换两个数组内容

Swap(int* e1, int* e2)

{

while (*e2)

{

int tmp = *e1;

*e1 = *e2;

*e2 = tmp;

e1++;

e2++;

}

}

//compal用来比较两个数组的大小

int compal(char* e1, char* e2)

{

if (strlen(e1) >strlen(e2))

return 0;

else

{

if (strlen(e1) < strlen(e2))

return 1;

else if (strlen(e1) == strlen(e2))

{

if (strcmp(e1, e2) < 0)

return 1;

else if (strcmp(e1, e2) > 0)

return 0;

}

}

}

int main()

{

char arr1[1000] = { "0" };

char arr2[1000] = { "0" };

int arr3[1001] = { 0 };

scanf("%s %s", arr1,arr2);

if (compal(arr1,arr2))

{

Swap(arr1,arr2);

int ch1[1000] = { 0 };

int ch2[1000] = { 0 };

int i = 0;

int len1 =strlen(arr1);

int len2 =strlen(arr2);

int len = len1 >len2 ? len1 : len2;

for (i = len1 - 1; i>= 0; i--)

{

ch1[i]= arr1[len1 - i - 1] - 48;

}

for (i = len2 - 1; i>= 0; i--)

{

ch2[i]= arr2[len2 - i - 1] - 48;

}

int arrlen = len + 1;

for (i = 0; i <arrlen; i++)

{

if (ch1[i] <ch2[i])

{

ch1[i]+= 10;

ch1[i+ 1] -= 1;

}

arr3[i]= ch1[i] - ch2[i];

}

while ((arr3[arrlen] == 0&& arrlen > 0))

arrlen--;

//printf("%d ", arrlen);

printf("-");

for (i = arrlen; i>= 0; i--)

{

printf("%d", arr3[i]);

}

}

else

{

int ch1[1000] = { 0 };

int ch2[1000] = { 0 };

int i = 0;

int len1 = strlen(arr1);

int len2 =strlen(arr2);

int len = len1 >len2 ? len1 : len2;

for (i = len1 - 1; i>= 0; i--)

{

ch1[i]= arr1[len1 - i - 1] - 48;

}

for (i = len2 - 1; i>= 0; i--)

{

ch2[i]= arr2[len2 - i - 1] - 48;

}

int arrlen = len + 1;

for (i = 0; i <arrlen; i++)

{

if (ch1[i] <ch2[i])

{

ch1[i]+= 10;

ch1[i+ 1] -= 1;

}

arr3[i]= ch1[i] - ch2[i];

}

while ((arr3[arrlen] == 0&& arrlen > 0))

arrlen--;

//printf("%d ", arrlen);

for (i = arrlen; i>= 0; i--)

printf("%d", arr3[i]);

}

return 0;

}

//乘法

int main()

{

char arr1[1000] = { "0" };

char arr2[1000] = { "0" };

int arr3[1001] = {0};

scanf("%s %s",arr1,arr2);

int ch1[1000] = {0};

int ch2[1000]={0};

int i = 0;

int len1 =strlen(arr1);

int len2 =strlen(arr2);

int arrlen = len1 +len2;

//int sub[1000][1000] = {0};

int len = len1 >len2 ? len1 : len2;

for (i = len1 ; i >0; i--)

{

ch1[i]= arr1[len1-i]-48;

}

for (i = len2; i >0; i--)

{

ch2[i]= arr2[len2-i]-48;

}

int j = 0;

for ( i= 1; i <len1+1; i++)

{

for (j = 1; j <len2+1; j++)

{

//核心算法

arr3[i+ j -1] += ch1[i]*ch2[j];

arr3[i+ j] += arr3[i + j - 1] / 10;

arr3[i+ j - 1] %= 10;

}

}

while ((arr3[arrlen] == 0&& arrlen >1))

arrlen--;

//printf("%d ", arrlen);

for (i = arrlen; i>0; i--)

printf("%d", arr3[i]);

return 0;

}