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 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 (<) 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 1, 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题意：
用科学计数法表示两个浮点数，判断保留n位小数的两个数字是否相等。
思路：
我们用cnta，cntb来表示两个浮点数的小数点的位置，pa， pb来表示两个数字第一个非0的位置。indexA[], indexB[] 用来表示结果的有效位数，expA， expB用来表示指数。
Code：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int main() { 6 int n; 7 string a, b, tempA, tempB; 8 cin >> n >> a >> b; 9 int expA, expB; 10 int cntA = a.length(), cntB = b.length(); 11 for (int i = 0; i < a.length(); ++i) { 12 if (a[i] == '.') { 13 cntA = i; 14 break; 15 } 16 } 17 for (int i = 0; i < b.length(); ++i) { 18 if (b[i] == '.') { 19 cntB = i; 20 break; 21 } 22 } 23 24 int pA = 0, pB = 0; 25 while (a[pA] == '0' || a[pA] == '.') pA++; 26 while (b[pB] == '0' || b[pB] == '.') pB++; 27 28 if (cntA < pA) 29 expA = cntA - pA + 1; 30 else 31 expA = cntA - pA; 32 if (pA == a.length()) expA = 0; 33 34 if (cntB < pB) 35 expB = cntB - pB + 1; 36 else 37 expB = cntB - pB; 38 if (pB == b.length()) expB = 0; 39 40 int indexA = 0, indexB = 0; 41 while (indexA < n) { 42 if (a[pA] != '.' && pA < a.length()) 43 tempA += a[pA++]; 44 else 45 tempA += '0'; 46 indexA++; 47 } 48 while (indexB < n) { 49 if (b[pB] != '.' && pB < b.length()) 50 tempB += b[pB++]; 51 else 52 tempB += '0'; 53 indexB++; 54 } 55 if (tempA == tempB && expA == expB) { 56 cout << "YES 0." << tempA << "*10^" << expA << endl; 57 } else { 58 cout << "NO 0." << tempA << "*10^" << expA << " 0." << tempB << "*10^" 59 << expB << endl; 60 } 61 return 0; 62 }
这种题难道是不难，就是细节太多，有一点考虑不到，就会被设计好的样例卡到。（上面的代码有一组数据没有通过）