• PAT甲级 科学计数法


    1073 Scientific Notation

    分数 20

    Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at least one digit in the fractional portion, and the number and its exponent's signs are always provided even when they are positive.

    Now given a real number A in scientific notation, you are supposed to print A in the conventional notation while keeping all the significant figures.

    Input Specification:

    Each input contains one test case. For each case, there is one line containing the real number A in scientific notation. The number is no more than 9999 bytes in length and the exponent's absolute value is no more than 9999.

    Output Specification:

    For each test case, print in one line the input number A in the conventional notation, with all the significant figures kept, including trailing zeros.

    Sample Input 1:

    +1.23400E-03
    

    Sample Output 1:

    0.00123400
    

    Sample Input 2:

    -1.2E+10
    

    Sample Output 2:

    -12000000000
    

    代码长度限制

    16 KB

    时间限制

    200 ms

    内存限制

    64 MB

    1. #include
    2. using namespace std;
    3. int main(){
    4. string origin;
    5. cin>>origin;
    6. char tag1,tag2;
    7. string kuo;
    8. int num;
    9. tag1=origin[0];
    10. if(tag1=='-') cout<<'-';
    11. int j=0;
    12. for(int i=origin.size()-1;i>=0;i--)if(origin[i]=='E') j=i;
    13. tag2=origin[j+1];
    14. kuo = origin.substr(j+2,origin.size());
    15. num = stoi(kuo);
    16. string now=origin.substr(1,j-1);
    17. if(tag2=='+'){
    18. int i=2;
    19. cout<0];
    20. while(num--){
    21. if(i>=now.size()) cout<<'0';
    22. else {
    23. cout<
    24. }
    25. i++;
    26. }
    27. if(now.size()>i){
    28. cout<<'.';
    29. while(isize()){
    30. cout<
    31. i++;
    32. }
    33. }
    34. }else{
    35. if(num==0)cout<
    36. else{
    37. cout<<"0.";
    38. num--;
    39. while(num--)cout<<'0';
    40. for(int i=0;isize();i++){
    41. if(now[i]!='.')cout<
    42. }
    43. }
    44. }
    45. }

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  • 原文地址:https://blog.csdn.net/qq_62368250/article/details/134562528