Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at least one digit in the fractional portion, and the number and its exponent’s signs are always provided even when they are positive.
Now given a real number A in scientific notation, you are supposed to print A in the conventional notation while keeping all the significant figures.
Input Specification:
Each input contains one test case. For each case, there is one line containing the real number A in scientific notation. The number is no more than 9999 bytes in length and the exponent’s absolute value is no more than 9999.
Output Specification:
For each test case, print in one line the input number A in the conventional notation, with all the significant figures kept, including trailing zeros.
Sample Input 1:
+1.23400E-03
Sample Output 1:
0.00123400
Sample Input 2:
-1.2E+10
Sample Output 2:
-12000000000
Code:
using namespace std;
//这题看到9999Bytes这个数量级，就一定用字符串存了
int main(int argc, char*argv[]) {
int zhishu;
char zhengfu, zhengshu[20000], xiaoshu[20000];//要开到两倍于9999B
char buffer[20000];
scanf("%s", buffer);
//itoa atoi?
//scanf("%c%s.%sE%d", &zhengfu, zhengshu, xiaoshu, &zhishu);
zhengfu = buffer[0];
int cursor = 1;
while (buffer[cursor] != '.') {
zhengshu[cursor - 1] = buffer[cursor];
cursor++;
}
zhengshu[cursor++ - 1] = '\0';
int len1 = strlen(zhengshu);
while (buffer[cursor] != 'E') {
xiaoshu[cursor - len1 - 2] = buffer[cursor];//这里数组访问到了xiaoshu[-1]，可以操作但会报错，Stack around the variable 'xiaoshu' was corrupted
cursor++;
}
xiaoshu[cursor++ - len1 - 2] = '\0';
int len2 = strlen(xiaoshu);
strcpy(buffer, buffer + cursor);//strncpy
zhishu = atoi(buffer);
//printf("%c%s.%sE%d\n", zhengfu, zhengshu, xiaoshu, zhishu);
if (zhengfu == '-')
printf("%c", zhengfu);
if (zhishu > 0) {//手动复制字符串
if (zhishu < len2) {
int i = 0, j = 0;
for (; i < zhishu; i++)
zhengshu[len1 + i] = xiaoshu[i];
zhengshu[len1 + i] = '\0';
for (; j < len2 - i; j++)
xiaoshu[j] = xiaoshu[j + i];
xiaoshu[j] = '\0';
printf("%s.%s", zhengshu, xiaoshu);
}
else {
int i = 0;
strcat(zhengshu, xiaoshu);
for (; i < zhishu - len2; i++)
buffer[i] = '0';
buffer[i] = '\0';
strcat(zhengshu, buffer);
printf("%s", zhengshu);
}
}
else if (zhishu < 0) {//用了字符串函数
zhishu = zhishu * (-1);
if (zhishu >= len1) {
int i = 0;
for (; i < zhishu - len1; i++)
buffer[i] = '0';
buffer[i] = '\0';
strcat(zhengshu, xiaoshu);
strcat(buffer, zhengshu);
strcpy(xiaoshu, buffer);
zhengshu[0] = '0';
zhengshu[1] = '\0';
}
else {
strcat(zhengshu, xiaoshu);
strcpy(xiaoshu, zhengshu + len1 - zhishu);
zhengshu[len1 - zhishu] = '\0';
}
printf("%s.%s", zhengshu, xiaoshu);
}
else {
printf("%s.%s", zhengshu, xiaoshu);
}
system("pause");
return 0;
}
//写得不够简洁，看这个https://www.liuchuo.net/archives/2061，用了string。
//或者看书上的也使用了char[]，都比我写得简单得多。