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POJ 3300 Tour de France(简单题)

栏目：互联网时间：2014-10-16 11:54:05

【题意简述】：由The drive ratio -- the ratio of the angular velocity of the pedals to that of the wheels -- isn : m where n is the number of teeth on the rear sprocket andm is the number of teeth on the front sprocket. Two drive ratios d₁< d₂ are adjacent if there is no other drive ratio d₁< d₃< d₂. The spread between a pair of drive ratiosd₁< d₂ is their quotient: d₂ ?d₁. You are to compute the maximum spread between two adjacent drive ratios achieved by a particular pair of front and rear clusters.我们可以知道，这个ratios等于n/m，然后d（从1……n）就是每一个ratios的值，现在让我们求最大的di/d(i-1)的值。

【分析】：有些题就是这样，只要我们理解了题意就可以很快的A掉，但如果不理解题意，就……

代码借鉴：http://www.cnblogs.com/eric-blog/archive/2011/06/01/2067441.html

//236K 16Ms #include<iostream> #include<cstdio> #include<algorithm> using namespace std; using namespace std; double ratio[100]; int f,r; double front[10],rear[10]; double max_ratio; int index; int main() { while(cin>>f,f) { cin>>r; for(int i=0; i<f; i++) cin>>front[i]; for(int i=0; i<r; i++) cin>>rear[i]; index=0; for(int i=0; i<r; i++) { for(int j=0; j<f; j++) { ratio[index++]=rear[i]/front[j]; } } sort(ratio,ratio+index); for(int i=0; i<index-1; i++) ratio[i]=ratio[i+1]/ratio[i]; max_ratio=0.0; for(int i=0; i<index-1; i++) { if(max_ratio<ratio[i]) max_ratio=ratio[i]; } printf("%.2f ",max_ratio); } return 0; }

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