# 1060 Are They Equal (25 分)（科学计数法）(1060 are they equal (25 points))-其他

## 1060 Are They Equal (25 分)（科学计数法）(1060 are they equal (25 points))

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.

3 12300 12358.9

YES 0.123*10^5

3 120 128

### Sample Output 2:

``````NO 0.120*10^3 0.128*10^3
``````

### 分析：

• cnta 和 cntb 通过扫描字符串得到小数点所在的下标（初始化cnta cntb为字符串长度，即下标为strlen(str））
• 考虑到可能前面有多余的零，用 p 和 q 通过扫描字符串使 p q 开始于第一个非0（且非小数点）处的下标
• 如果cnta >= p ，说明小数点在第一个开始的非0数的下标的右边，那么科学计数法的指数为cnta – p ; 否则应该为cnta – p + 1; 字符串b同理。
• 如果 p 和 q 等于字符串长度， 说明字符串是 0， 此时直接把 cnta（或者cntb）置为0，因为对于0来说乘以几次方都是相等的，如果不置为0可能会出现两个0比较导致判断为它们不相等
• 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
• 判断A和B是否相等，且cnta和cntb是否相等。如果相等，说明他们用科学计数法表示后是相同的，输出YES，否则输出NO，同时输出正确的科学计数法

### 注意：

• 10的0次方和1次方都要写。
• 题目中说，无需四舍五入。
• 数组开大点，虽然只有100位，但是很有可能前面的0很多导致根本不止100位。一开始开的110，几乎没有AC的任何测试点。。后来开了10000就AC了~

### 柳神题解

https://blog.csdn.net/qq_41581765/article/details/119791168

``````#include <iostream>
#include <cstring>
using namespace std;
int main() {
int n, p = 0, q = 0;
char a[10000], b[10000], A[10000], B[10000];
scanf("%d%s%s", &n, a, b);
int cnta = strlen(a), cntb = strlen(b);
for(int i = 0; i < strlen(a); i++) {
if(a[i] == '.') {
cnta = i;
break;
}
}
for(int i = 0; i < strlen(b); i++) {
if(b[i] == '.') {
cntb = i;
break;
}
}
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 == strlen(a))
cnta = 0;
if(q == strlen(b))
cntb = 0;
while(indexa < n) {
if(a[p] != '.' && p < strlen(a))
A[indexa++] = a[p];
else if(p >= strlen(a))
A[indexa++] = '0';
p++;
}
while(indexb < n) {
if(b[q] != '.' && q < strlen(b))
B[indexb++] = b[q];
else if(q >= strlen(b))
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;
}
``````
————————

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.

3 12300 12358.9

YES 0.123*10^5

3 120 128

### Sample Output 2:

``````NO 0.120*10^3 0.128*10^3
``````

### Main idea of the title:

Give two numbers and ask whether they are equal after they are written into a scientific counting method with N decimal places. If equal, output yes and output their scientific notation representation. If no is not output equally, their scientific counting method is output respectively

### analysis:

• CNTA and cntb obtain the subscript of the decimal point by scanning the string (initial CNTA and cntb is the length of the string, that is, the subscript is strlen (STR))
• Considering that there may be redundant zeros in front of P and Q, scan the string with P and Q to make P and Q start at the subscript at the first non-0 (and non decimal point)
• If CNTA > = P, it means that the decimal point is to the right of the subscript of the first non-zero number, then the index of scientific counting method is CNTA – p; Otherwise, it should be CNTA – P + 1; The same is true for string B.
• If P and Q are equal to the length of the string, it means that the string is 0. At this time, CNTA (or cntb) is directly set to 0, because for 0, multiplying by several powers is equal. If it is not set to 0, two zeros may be compared, resulting in judgment that they are not equal
• Indexa = 0 starts to assign a value to the new a array, with a total of N normal digits excluding the decimal point, starting from the subscript of P. If P is greater than or equal to strlen, it means that the required number of digits is still not met after the traversal of the string. At this time, 0 needs to be added after array a until n digits are met. Similarly, indexb generates a new B array
• Judge whether a and B are equal and whether CNTA and cntb are equal. If they are equal, it means that they are the same after being expressed by scientific counting method, and output yes; otherwise, output no, and output the correct scientific counting method at the same time

### be careful:

• Write to the power 0 and power 1 of 10.
• The title says that there is no need to round.
• The array is a little larger. Although it is only 100 bits, it is likely that many previous zeros will lead to more than 100 bits at all. At the beginning of 110, there were almost no AC test points.. Then I drove 10000 to AC~

### Liu Shen’s explanation

A subject of scientific counting in class B
https://blog.csdn.net/qq_ 41581765/article/details/119791168
Comparative learning~

``````#include <iostream>
#include <cstring>
using namespace std;
int main() {
int n, p = 0, q = 0;
char a[10000], b[10000], A[10000], B[10000];
scanf("%d%s%s", &n, a, b);
int cnta = strlen(a), cntb = strlen(b);
for(int i = 0; i < strlen(a); i++) {
if(a[i] == '.') {
cnta = i;
break;
}
}
for(int i = 0; i < strlen(b); i++) {
if(b[i] == '.') {
cntb = i;
break;
}
}
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 == strlen(a))
cnta = 0;
if(q == strlen(b))
cntb = 0;
while(indexa < n) {
if(a[p] != '.' && p < strlen(a))
A[indexa++] = a[p];
else if(p >= strlen(a))
A[indexa++] = '0';
p++;
}
while(indexb < n) {
if(b[q] != '.' && q < strlen(b))
B[indexb++] = b[q];
else if(q >= strlen(b))
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;
}
``````