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# 1060 Are They Equal (25 分)（科学计数法）

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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：

1. 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））
2. 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
3. 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。
4. 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
5. 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
6. 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~

chopping 碎片

# 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;
}
``````