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
Meaning of the questions:
Given two numbers, Q: write them retained N (<100) decimal scientific notation 0.d[1]...d[N]*10^k ( d[1]> 0 IS 0 The unless The Number); after, are the same. If equal, the output YES, and the conversion result; if not equal, the NO output, and gives the conversion of two numbers.
answer:
The idea of this problem and the situation is difficult to think very complex judgment.
When the subject of the request scientific notation, the two numbers are equal. Thus when the body portion and exponent portion as long as the determination is equal to LCyT.
For data, it should be divided into less than 1 and greater than 1 to determine, to be considered in each case number, for example: 0000, 000.00, 00123.4, 0.012, 0.00, etc.
Initially did not consider too many exceptions, took 19 minutes, the next day saw the attention point solution to a problem redo, or only 21 points,
On the third day Lile idea, in fact, as long as the focus, the decimal points after a good format where the number is the number on it, on this basis, special sentenced to 0 to remove leading zeros. Compare careful not to overrun when the output is not 0 padded.
Own test sample:
2 0.0015 0000.001520000 YES 0.15*10^-2
3 010.25 0010.23 YES 0.102*10^2
5 0.1 00.100 YES 0.10000*10^0
10 000.0012 0000000.0012000000000 YES 0.1200000000*10^-2
3 0 0.000 YES 0.000*10^0
AC Code:
#include<iostream> #include<string> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int n; char a[105]; char b[105]; int main(){ cin>>n>>a>>b; int la=strlen(a); int lb=strlen(b); int sta=-1,stb=-1; int isa0=0,isb0=0; // Japanese sentence is not 0 // starting from the first number is not zero for ( int I = 0 ; I <La; I ++ ) { IF ! (A [I] = ' . ' && A [I] =! ' 0 ' ) { STA = I; BREAK ; } } for ( int I = 0 ; I <LB; I ++ ) { IF ! (B [I] = ' . ' ! && B [I] = ' 0 ' ) { STB = I; break; } } if(sta==-1) isa0=1; if(stb==-1) isb0=1; //小数点前有几位 int ia=0,ib=0; int f=0; for(int i=0;i<la;i++){ if(a[i]=='.') break; if(a[i]!='0') f=1; if(f){ ia++; } } f=0; for(int i=0;i<lb;i++){ if(b[i]=='.') break; if(b[i]!='0') f=1; if(f){ ib++; } } //cout<<"sta:"<<sta<<" ia:"<<ia<<" ha:"<<ha<<endl; //cout<<"stb:"<<stb<<" ib:"<<ib<<" hb:"<<hb<<endl; //比较 f=1; int pa=sta,pb=stb; if(isa0!=isb0 || ia!=ib) f=0; for(int i=0;i<n;i++){ if(pa+i>=la||pb+i>=lb) break; if(a[pa]=='.') pa++; if(b[pb]=='.') pb++; //cout<<"pa "<<pa+i<<" a[pa] "<<a[pa+i]<<endl; //cout<<"pb "<<pa+i<<" b[pb] "<<b[pa+i]<<endl; if(a[pa+i]!=b[pb+i]){ f=0; break; } } if(isa0==1&&isa0==isb0) f=1;//如果两个都是0 //输出 if(f){ cout<<"YES 0."; if(isa0==1){//0的特判 for(int i=1;i<=n;i++) cout<<"0"; cout<<"*10^0"; }else{ pa=sta; for(int i=0;i<n;i++){ if(pa+i>=la) { cout<<"0"; continue; } if(a[pa+i]=='.') pa++; if(pa+i>=la) cout<<"0"; else cout<<a[pa+i]; } cout<<"*10^"; if(ia!=0) cout<<ia;//次数>=于0 else{//次数<0 int k=-1; for(int i=0;i<la;i++){ if(a[i]=='.'){ k=i; break; } } COUT << K-STA + . 1 ; // . Position - the start position + 1 } } } the else { // supra COUT << " NO of 0. The " ; IF (ISA0 == . 1 ) { for ( int I = . 1 ; I <= n-; I ++) COUT << " 0 " ; COUT << " * 10 ^ 0 " ; } the else { for ( int I = 0 ; I <n-;i++){ if(pa+i>=la) { cout<<"0"; continue; } if(a[pa+i]=='.') pa++; if(pa+i>=la) cout<<"0"; else cout<<a[pa+i]; } cout<<"*10^"; if(ia!=0) cout<<ia<<" 0."; else{ int k=-1; for(int i=0;i<la;i++){ if(a[i]=='.'){ k=i; break; } } cout<<k-sta+1<<" 0."; } } if(isb0==1){ for(int i=1;i<=n;i++) cout<<"0"; cout<<"*10^0"; }else{ for(int i=0;i<n;i++){ if(pb+i>=lb) { cout<<"0"; continue; } if(b[pb+i]=='.') pb++; if(pb+i>=lb) cout<<"0"; else cout<<b[pb+i]; } cout<<"*10^"; if(ib!=0) cout<<ib; else{ int k=-1; for(int i=0;i<lb;i++){ if(b[i]=='.'){ k=i; break; } } cout<<k-stb+1; } } } return 0; }