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Hangover
POJ - 1003 Hangover
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) |
方法-: #include<stdio.h> int main(){ float x; while(scanf("%f",&x)==1){ int i; float t=0.0; if(x==0.00) break; for(i=1;t<x;i++) t+=1.0/(float)(i+1); printf("%d card(s)\n",i-1); } } 方法二: #include<stdio.h> int main(){ int n; float b; for(;n=scanf("%f",&b),b;printf("%d card(s)\n",n-1)) for(;b>0;b-=1./++n); }