1060 Are They Equal （附测试点）

1060 Are They Equal (25 分)

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 0.123×10​5​​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 Bare the two float numbers to be compared. Each float number is non-negative, no greater than 10​100​​, 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.d[1]...d[N]*10^k(d[1]>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

看似不难的一道题，思路很清晰，就是转换成规定的科学计数法之后比较

测试数据却是十分孤儿，最后符测试点

#include<bits/stdc++.h> using namespace std; string s1, s2, a1, a2; string translate(int n, string s) { int size = s.size(), i = 0, count = 0; string temp; for (; i < size&&s[i] == '0'; i++); if (s[i] == '.') { i++; for (; s[i] == '0'; i++)count--; bool judge = false; //防止0.0这样的数据，只有judge为true时，count才有效 for (; temp.size() < n&&i < size; i++) { temp.push_back(s[i]); judge = true; } while (temp.size() < n)temp.push_back('0'); if(!judge)count=0; } else { for(int j=i;s[j]!='.'&&j<size;j++)count++; for (; s[i] != '.'&&temp.size()<n&&i<size; i++) temp.push_back(s[i]); if (temp.size() < n)i++; for (; i < size&&temp.size() < n; i++) temp.push_back(s[i]); while (temp.size() < n)temp.push_back('0'); } temp += ("*10^" + to_string(count)); return temp; } int main() { int n = 0; scanf("%d", &n); cin >> s1 >> s2; a1 = translate(n, s1); a2 = translate(n, s2); if (a1 == a2) cout << "YES 0." << a1; else cout << "NO 0." << a1 << " 0." << a2; return 0; }

测试点：

 3 12300 12358.9 YES 0.123*10^5 1 12300 12358.9 YES 0.1*10^5 1 1.2300 1.23589 YES 0.1*10^1 5 1.2300 1.23589 NO 0.12300*10^1 0.12358*10^1 4 0.01234 0.012345 YES 0.1234*10^-1 5 0.01234 0.012345 NO 0.12340*10^-1 0.12345*10^-1 5 0.1234 0.12345 NO 0.12340*10^0 0.12345*10^0 0 0.11 0 YES 0.*10^0或者YES 0.0*10^0，都可以AC，测试点应该没有这个例子 1 0.1 0 NO 0.1*10^0 0.0*10^0 1 0.0 0.1 NO 0.0*10^0 0.1*10^0 1 0.0 0.000 YES 0.0*10^0 1 00.0 0.000 YES 0.0*10^0 4 00.0 0.000 YES 0.0000*10^0 5 00.0 0.000 YES 0.00000*10^0 1 05.0 5.000 YES 0.5*10^1 1 00.01 0.010 YES 0.1*10^-1