HangOver
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13654 Accepted Submission(s): 5966
Problem 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.
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.
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.
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.
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)
水题
//
// Created by Admin on 2017/4/4.
//
#include <cstdio>
int main(){
double a[1010],sum=0.0,temp;
for (int i = 1; i < 300; ++i) {
a[i]=1.0/(i+1)+sum;
sum+=1.0/(i+1);
}
//for (int i = 1; i < 300; ++i) printf("%.2f\n",a[i]);
while (scanf("%lf",&temp),temp){
for (int i = 1; i < 300; ++i) {
if(a[i]>=temp){
printf("%d card(s)\n",i);
break;
}
}
}
return 0;
}