1060. Are They Equal (25)-PAT甲级真题

1060.Are They Equal (25)

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line “YES” if the two numbers are treated equal, and then the number in the standard form “0.d1…dN*10^k” (d1>0 unless the number is 0); or “NO” if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

　　题目大意：给出两个数，问将它们写成保留N位小数的科学计数法后是否相等。如果相等，输出YES，输出他们的科学记数法表示方法。如果不相等输出NO，分别输出他们的科学计数法。

　　

#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int main ()
{
int n, p = 0, q = 0;
string a, b;
char A[10000], B[10000];
scanf ("%d%s%s", &n, a, b);\
/*
cnta,cntb是用来记录科学技术法中10的次数的，
因为按照题目里最后都是0.xxx，
所以就是把所有的位数全部移动到小数点的右边，
也就是计算小数点左边到底有几个数。
那么就先找到小数点的index，
如果出现0和.计数位p,q++,出现其他数字就结束循环，
这样可以划分出整数位是0和整数位不是0
*/
int cnta = a.find ('.');
int cntb = b.find ('.');
int indexa = 0, indexb = 0;
while (a[p] == '0' || a[p] == '.')
p++;
while (b[q] == '0' || b[q] == '.')
q++;
if (cnta >= p)
cnta = cnta - p;
else
cnta = cnta - p + 1;
if (cntb >= q)
cntb = cntb - q;
else
cntb = cntb - q + 1;
if (p == a.length ())
cnta = 0;
if (q == b.length())
cntb = 0;
/*
indexa = 0开始给新的A数组赋值，共赋值n位除去小数点外的正常数字，
从p的下标开始。如果p大于等于strlen，
说明字符串遍历完毕后依旧没能满足需要的位数，
此时需要在A数组后面补上0直到满足n位数字。
indexb同理，产生新的B数组
*/
while (indexa < n)
{
if (a[p] != '.' && p < a.length ())
A[indexa++] = a[p];
else if (p >= a.length ())
A[indexa++] = '0';
p++;
}
while (indexb < n)
{
if (b[q] != '.' && q < b.length ())
B[indexb++] = b[q];
else if (q >= b.length())
B[indexb++] = '0';
q++;
}
if (strcmp (A, B) == 0 && cnta == cntb)
printf ("YES 0.%s*10^%d", A, cnta);
else
printf ("NO 0.%s*10^%d 0.%s*10^%d", A, cnta, B, cntb);
return 0;
}