POJ 1003(Hangover 简单数学) 解题报告

/*___________________________________________POJ 1002题_________________________________________________________ Hangover Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 56408 Accepted: 26444 Description: How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below. Input: The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits. Output: For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples. Sample Input: 1.00 3.71 0.04 5.19 0.00 Sample Output: 3 card(s) 61 card(s) 1 card(s) 273 card(s) **********************************************************************************************************/ #include<iostream> #include<string> using namespace std; //* #include<fstream> ifstream fin("input.txt"); #define cin fin //*/ string Calculate(int n) { char str[10]; float i; float result=0; for(i=2;i<=n+1;i++) result+=1/i; //cout<<result<<endl; sprintf(str,"%5.3f",result);//先留出3位小数 //cout<<str<<endl; string temp(str); temp.erase(temp.length()-1,1);//把最后一位小数直接删除，而不是四舍五入 return temp; } int main() { int n; char str[10]; float length; // cout<<Calculate(272)<<endl; // cout<<Calculate(273)<<endl; string temp; cin>>length; //cout<<length<<endl; sprintf(str,"%4.2f",length); temp=str; //cout<<temp<<endl; do { for(n=1;n<500;n++) { //cout<<n<<" "<<Calculate(n)<<endl; if(temp>Calculate(n-1) && temp<=Calculate(n)) { cout<<n<<" card(s)"<<endl; break; } } cin>>length; //cout<<length<<endl; sprintf(str,"%4.2f",length); temp=str; //cout<<temp<<endl; }while(temp!="0.00"); return 0; } //总结：注意精度，舍入