POJ 1003 Hangover
摘要:        
POJ 1003 Hangover 水题//POJ 1003#include <iostream>using namespace std;float l[1001];int main(){ float f; l[0] = 0; for (int i = 1; i <= 1000; i++) l[i] = l[i-1]+1.0/(i+1); while (cin>>f, f!=0.00) { for (int i = 1; i <= 1000; i++) if (l[i] > f) { ...    阅读全文
		
		posted @ 2012-05-25 11:44 澄哥 阅读(314) 评论(0) 推荐(0)
                    
                
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