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 B are 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
题目大意：给定有效位数， 判断两个浮点数是否相同， 并且输出浮点表示
思路：找到第一个非0数字的位置a0，以及小数点的位置adot, 如果a0>adot,则指数expa=a0-adot, 否则expa=a0-adot+1。删除小数点，以及有效位之前的0； 根据有效数字增加0，或者截断数字
注意点：当数字是0的时候指数一定的0，不管小数点在什么位置。 还有就是除了数字相同，还要指数相同的两个数字才是相等的。 开始没有考虑指数是负数需要加1，导致有一个点不能通过
1 #include<iostream> 2 #include<string> 3 using namespace std; 4 int main(){ 5 int n, i; 6 string a, b; 7 cin>>n>>a>>b; 8 int la=a.size(), lb=b.size(), adot=la, bdot=lb, a0=0, b0=0; 9 for(i=0; i<la; i++) if(a[i]=='.'){adot=i; break;} 10 for(i=0; i<lb; i++) if(b[i]=='.'){bdot=i; break;} 11 for(i=0; i<la; i++) if(a[i]>='1'&&a[i]<='9'){a0=i; break;} 12 for(i=0; i<lb; i++) if(b[i]>='1'&&b[i]<='9'){b0=i; break;} 13 int expa=adot-a0, expb=bdot-b0; 14 a.erase(adot,1); b.erase(bdot, 1); 15 for(i=0; i<a.size(); i++) if(a[i]!='0'){a0=i; break;} 16 for(i=0; i<b.size(); i++) if(b[i]!='0'){b0=i; break;} 17 a.erase(0, a0); b.erase(0, b0); 18 la=a.size(); lb=b.size(); 19 for(i=0; i<n-la; i++) a.append("0"); 20 for(i=0; i<n-lb; i++) b.append("0"); 21 a.erase(n); b.erase(n); 22 if(a[0]=='0') expa=0; 23 if(b[0]=='0') expb=0; 24 if(expa<0) expa++; 25 if(expb<0) expb++; 26 if(expa==expb && a==b) cout<<"YES 0."<<a<<"*10^"<<expa<<endl; 27 else cout<<"NO 0."<<a<<"*10^"<<expa<<" 0."<<b<<"*10^"<<expb<<endl; 28 return 0; 29 }