hdu Pie

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这道题是一道二分搜索的题,首先计算出最大的平均体积:mx=V总/f:然后去left=0,right=mx,mid=(left+right)/2进行二分搜索,当所有pi分割出的mid的个数是大于等于f时,说明这个mid是满足条件的,一直搜索知道找到最大的mid:这题的核心还是精度的问题.不得不说这题对精度的要求实在是太高了.首先是mi的精度,当我用的是mi=1e-8时,会超时,慢慢将mi降到1e-6.后来又发现当将pi定义为3.1415926535是个会wa掉,后来参考了一下别人的代码才知道原来d…

hdu 1969 Pie（二分查找）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1969 Pie Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4513 Accepted Submission(s): 1819 Problem Description My birthday is coming up and trad…

HDU 1969 Pie（二分法）

My birthday is coming up and traditionally I’m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece…

HDU 3392 Pie(DP)

题意:有一些男生女生,男生女生数量差不超过100 ,男生女生两两配对.要求求出一种配对方法,使每一对的高度差的和最小. 思路:(我是真的笨笨笨!!)设人少的一组人数为n,b[],人多的一组人数为m,g[](b[],g[]先排好序),用dp[i][j]表示n中的前i个人与m中的前j个人配对所得到的最小值. 对于每个i,j的范围是(i~i+m-n) 这样j的范围是i~m-n+i,对于dp数组而言,第二维数组要开到m,由于数组过大,而我们又知道,m-n<=100,所以用j表示j-i,j的范围就变成了(…

Pie（hdu 1969 二分查找）

Pie Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7302 Accepted Submission(s): 2732 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I h…

HDU 1969 Pie(二分查找)

Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This…

HDU 1969 Pie(二分搜索)

题目链接 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie.…