HangOver
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7693????Accepted Submission(s): 3129
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.
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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.
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Sample Input
1.00 3.71 0.04 5.19 0.00
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Sample Output
3 card(s) 61 card(s) 1 card(s) 273 card(s)
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Source
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題目沒什么難度,分明就是某年NOIP的級數(shù)求和,不過題目里如果不說,我還真不一定能想到,這個(gè)結(jié)論得記一下.
#include<stdio.h> #include < string .h> int f[ 1024 ]; double s[ 1024 ]; void getprepared() { memset(f, 0 , sizeof (f)); memset(s, 0 , sizeof (s)); s[ 1 ]= 0.5 ; for ( int i= 2 ;i<= 1000 ;i++) s[i]=s[i- 1 ]+ 1.0 /(i+ 1 ); for ( int i= 1 ;i<= 520 ;i++ ) { double x=i/ 100.0 ; for ( int j= 1 ;j<= 1000 ;j++ ) if (s[j]>= x) { f[i] = j; break ; } } } int main() { getprepared(); double ss; while (scanf( " %lf " ,&ss)!= EOF) { if (ss== 0 ) return 0 ; int x= 100 * ss; printf( " %d card(s)\n " ,f[x]); } return 0 ; }
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