題目鏈接：http://codeforces.com/contest/697/PRoblem/B
【中文題意】給你一個(gè)浮點(diǎn)數(shù)的計(jì)數(shù)法表示形式，然后表示出他的正常形式，表示出的數(shù)不能有前導(dǎo)0也不能有后導(dǎo)0。 【思路分析】直接模擬就好了，注意幾個(gè)坑：1. 0.0e0的時(shí)候 2. 1.0e0的時(shí)候。
//第一個(gè)是本弱雞的代碼，寫了100多行 【AC代碼】
#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<queue>#include<stack>#include<map>#include<algorithm>using namespace std;#define LL long longint main(){ string str; while(cin>>str) { int len=str.length(); int dightPosition,ePosition; for(int i=0;i<len;i++) { if(str[i]=='.') { dightPosition=i; } if(str[i]=='e') { ePosition=i; } } int num1=0,num2=0; for(int i=0;i<dightPosition;i++) { num1=num1*10+str[i]-'0'; } for(int i=dightPosition+1;i<ePosition;i++) { num2=num2*10+str[i]-'0'; } if(num1==0&&num2==0) { cout<<"0"<<endl; continue; } int square=0; for(int i=ePosition+1;i<len;i++) { square=square*10+str[i]-'0'; } //cout<<dightPosition<<ePosition<<square<<endl; if(square==0) { if(num2==0) { for(int i=0;i<dightPosition;i++) { cout<<str[i]; } } else { for(int i=0;i<ePosition;i++) { cout<<str[i]; } } cout<<endl; continue; } string str2; for(int i=0;i<dightPosition;i++) { str2+=str[i]; } if(dightPosition+square==ePosition-1) { //cout<<"1."<<endl; for(int i=dightPosition+1;i<ePosition;i++) { str2+=str[i]; } //cout<<str2<<endl; } if(dightPosition+square<ePosition-1) { //cout<<"2."<<endl; for(int i=dightPosition+1;i<dightPosition+square+1;i++) { str2+=str[i]; } str2+='.'; for(int i=dightPosition+square+1;i<ePosition;i++) { str2+=str[i]; } } if(dightPosition+square>ePosition-1) { //cout<<"3."<<endl; for(int i=dightPosition+1;i<ePosition;i++) { str2+=str[i]; } for(int i=0;i<square-ePosition+dightPosition+1;i++) { str2+='0'; } } //str2+='/0'; int len2=str2.length(),flag=0; //cout<<str2<<endl; for(int i=0;i<len2;i++) { if(str2[i]=='0'&&str2[i+1]=='.') { flag=1; } if(str2[i]=='0'&&flag==1) { cout<<str2[i]; } if(str2[i]!='0') { cout<<str2[i];flag=1; } } cout<<endl; } return 0;}//大佬的java代碼
import java.util.*;import java.math.*;public class B { public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.nextBigDecimal().stripTrailingZeros().toPlainString(); if (s.endsWith(".0")) { s = s.substring(0, s.length() - 2); } System.out.println(s); }}//大佬的C++代碼
#include<bits/stdc++.h>using namespace std;int ma(int a, int b){ return (a>b?a:b);}int main(){ string s; cin >> s; string round; int aftere = 0; int eloc = s.length(); int i,j,k,l; for(i = 0; i < s.length();i++) { if(s[i] == 'e') eloc = i; if(i > eloc) { aftere*=10; aftere += (s[i]-'0'); } } round.resize(eloc-1); j = 0; int dotloc; for(i = 0; i < eloc;i++) { if(s[i]!='.') { round[j] = s[i]; j++; } else dotloc = j; } dotloc += aftere; bool all0 = true; for(i = dotloc; i < round.length();i++) { if(round[i] != '0') all0 = false; } for(i = 0; i < ma(round.length(),dotloc);i++) { if(all0 && i == dotloc) return 0; if(i == dotloc) cout << '.'; if(i < round.length()) cout << round[i]; else cout << 0; }}新聞熱點(diǎn)
疑難解答
圖片精選
網(wǎng)友關(guān)注
