大整数加法 C语言 求代码 要求如下
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发布时间：2022-04-28 12:08
共1个回答
热心网友
时间：2023-10-08 22:53
#include<string.h>
void reverse( char *s ) /*将字符串逆置*/
{
int length;
int i = 0;
char temp;
length = strlen( s );
while( i < length - i - 1 )
{
temp = s[i];
s[i] = s[length - i - 1];
s[length - i - 1] = temp;
i++;
}
}
void AddBigNum( char* s1, char* s2, char* result )
{
int len1 = strlen( s1 );
int len2 = strlen( s2 );
int acc = 0, temp, i; /*acc为进位标记*/
if( s1 == NULL || s2 == NULL || result == NULL )
{
return;
}
reverse( s1 );
reverse( s2 );
for( i = 0; i < len1 && i < len2; i++ )
{
temp = s1[i] - '0' + s2[i] - '0' + acc; /*计算每位的实际和*/
result[i] = temp % 10 + '0'; /*通过求余数来确定每位的最终值*/
if( temp >= 10 ) /*通过这个if..else..条件来判断是否有进位，并设置进位值*/
acc = 1;
else
acc = 0;
}
if( i < len1 ) /*两个加数位数不同*/
{
for( ; i < len1; i++ )
{
temp = s1[i] - '0' + acc; /*依旧要考虑进位，比如9999 + 1的情况*/
result[i] = temp % 10 + '0';
if( temp >= 10 )
acc = 1;
else
acc = 0;
}
}
if( i < len2 )
{
for( ; i < len2; i++ )
{
temp = s2[i] - '0' + acc;
result[i] = temp % 10 + '0';
if( temp >= 10 )
acc = 1;
else
acc = 0;
}
}
if( acc == 1 ) /*考虑如:123 + 911 = 1034的情况，如果不增加这个条件会得到结果为034，进位被舍弃*/
result[i++] = '1';
result[i] = '\0';
reverse( result );
}
main()
{
char s1[405];
char s2[405];
char result[405];
int N;
scanf("%d",&N);
while(N--)
{
scanf("%s%s",s1,s2);
AddBigNum( s1, s2, result );
printf( "%s\n", result );
}
return 0;
}
运行结果如图：
热心网友
时间：2023-10-08 22:53
#include<string.h>
void reverse( char *s ) /*将字符串逆置*/
{
int length;
int i = 0;
char temp;
length = strlen( s );
while( i < length - i - 1 )
{
temp = s[i];
s[i] = s[length - i - 1];
s[length - i - 1] = temp;
i++;
}
}
void AddBigNum( char* s1, char* s2, char* result )
{
int len1 = strlen( s1 );
int len2 = strlen( s2 );
int acc = 0, temp, i; /*acc为进位标记*/
if( s1 == NULL || s2 == NULL || result == NULL )
{
return;
}
reverse( s1 );
reverse( s2 );
for( i = 0; i < len1 && i < len2; i++ )
{
temp = s1[i] - '0' + s2[i] - '0' + acc; /*计算每位的实际和*/
result[i] = temp % 10 + '0'; /*通过求余数来确定每位的最终值*/
if( temp >= 10 ) /*通过这个if..else..条件来判断是否有进位，并设置进位值*/
acc = 1;
else
acc = 0;
}
if( i < len1 ) /*两个加数位数不同*/
{
for( ; i < len1; i++ )
{
temp = s1[i] - '0' + acc; /*依旧要考虑进位，比如9999 + 1的情况*/
result[i] = temp % 10 + '0';
if( temp >= 10 )
acc = 1;
else
acc = 0;
}
}
if( i < len2 )
{
for( ; i < len2; i++ )
{
temp = s2[i] - '0' + acc;
result[i] = temp % 10 + '0';
if( temp >= 10 )
acc = 1;
else
acc = 0;
}
}
if( acc == 1 ) /*考虑如:123 + 911 = 1034的情况，如果不增加这个条件会得到结果为034，进位被舍弃*/
result[i++] = '1';
result[i] = '\0';
reverse( result );
}
main()
{
char s1[405];
char s2[405];
char result[405];
int N;
scanf("%d",&N);
while(N--)
{
scanf("%s%s",s1,s2);
AddBigNum( s1, s2, result );
printf( "%s\n", result );
}
return 0;
}
运行结果如图：
热心网友
时间：2023-10-08 22:53
#include<string.h>
void reverse( char *s ) /*将字符串逆置*/
{
int length;
int i = 0;
char temp;
length = strlen( s );
while( i < length - i - 1 )
{
temp = s[i];
s[i] = s[length - i - 1];
s[length - i - 1] = temp;
i++;
}
}
void AddBigNum( char* s1, char* s2, char* result )
{
int len1 = strlen( s1 );
int len2 = strlen( s2 );
int acc = 0, temp, i; /*acc为进位标记*/
if( s1 == NULL || s2 == NULL || result == NULL )
{
return;
}
reverse( s1 );
reverse( s2 );
for( i = 0; i < len1 && i < len2; i++ )
{
temp = s1[i] - '0' + s2[i] - '0' + acc; /*计算每位的实际和*/
result[i] = temp % 10 + '0'; /*通过求余数来确定每位的最终值*/
if( temp >= 10 ) /*通过这个if..else..条件来判断是否有进位，并设置进位值*/
acc = 1;
else
acc = 0;
}
if( i < len1 ) /*两个加数位数不同*/
{
for( ; i < len1; i++ )
{
temp = s1[i] - '0' + acc; /*依旧要考虑进位，比如9999 + 1的情况*/
result[i] = temp % 10 + '0';
if( temp >= 10 )
acc = 1;
else
acc = 0;
}
}
if( i < len2 )
{
for( ; i < len2; i++ )
{
temp = s2[i] - '0' + acc;
result[i] = temp % 10 + '0';
if( temp >= 10 )
acc = 1;
else
acc = 0;
}
}
if( acc == 1 ) /*考虑如:123 + 911 = 1034的情况，如果不增加这个条件会得到结果为034，进位被舍弃*/
result[i++] = '1';
result[i] = '\0';
reverse( result );
}
main()
{
char s1[405];
char s2[405];
char result[405];
int N;
scanf("%d",&N);
while(N--)
{
scanf("%s%s",s1,s2);
AddBigNum( s1, s2, result );
printf( "%s\n", result );
}
return 0;
}
运行结果如图：
