POJ 1759 Garland  这个题wa了27次,忘了用一个数来储存f[n-1],每次由于二分都会改变f[n-1]的值,得到的有的值不精确,直接输出f[n-1]肯定有问题. 这个题用c++交可以过,g++交过不了, f[i]=2+2*f[i-1]-f[i-2]; f[0]=A,f[1]=x; 二分枚举x的值,最终得到B的值(不是f[n-1]), 上述递推式是非齐次的二阶递推式,解其齐次特征方程,可以得到f[n-1]的齐次时的表达式,对于非齐次的,目前我还没法求出通式, B与f[n-1]的关…

Garland Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2365   Accepted: 1007 Description The New Year garland consists of N lamps attached to a common wire that hangs down on the ends to which outermost lamps are affixed. The wire sags…

题目链接:http://poj.org/problem?id=3104                                                                                                      Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11128   Accepted: 2865 Description It is very…

题意:给你n个派,每个派都是高为一的圆柱体,把它等分成f份,每份的最大体积是多少. 思路: 明显的二分答案题-- 注意π的取值- 3.14159265359 这样才能AC,,, //By SiriusRen #include <cstdio> using namespace std; int n,f,cases,a[10050]; double v[10050]; bool check(double x){ int ans=0; for(int i=1;i<=n;i++)ans+=v[i…

[题目链接] http://poj.org/problem?id=1759 [题目大意] 有n个数字H,H[i]=(H[i-1]+H[i+1])/2-1,已知H[1],求最大H[n], 使得所有的H均大于0. [题解] 我们得到递推式子H[i]=2*H[i-1]+2-H[i-2],发现H[n]和H[2]成正相关 所以我们只要二分H[2]的取值,同时计算每个H是否大于等于0即可. [代码] #include <cstdio> int n; double H[1010],A,B; bool che…

Description The New Year garland consists of N lamps attached to a common wire that hangs down on the ends to which outermost lamps are affixed. The wire sags under the weight of lamp millimeter lower than the average height of the two adjacent lamps…

睡觉啦 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; ,M=2e6+,INF=1e7; inline int read(){ ,f=; ;c=getchar();} +c-';c=getchar();} return x…

Musical Theme Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 28435 Accepted: 9604 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the pian…

Long Long Message Problem's Link:http://poj.org/problem?id=2774 Mean: 求两个字符串的最长公共子串的长度. analyse: 前面在学习后缀数组的时候已经做过一遍了,但是现在主攻字符串hash,再用字符串hash写一遍. 这题的思路是这样的: 1)取较短的串的长度作为high,然后二分答案(每次判断长度为mid=(low+high)>>1是否存在,如果存在就增加下界:不存在就缩小上界): 2)主要是对答案的判断(judge函数…

题目链接:http://poj.org/problem?id=1064 有n条绳子,长度分别是Li.问你要是从中切出m条长度相同的绳子,问你这m条绳子每条最长是多少. 二分答案,尤其注意精度问题.我觉得关于浮点数的二分for循环比while循环更好一点.注意最后要用到floor 保证最后答案不会四舍五入. #include <iostream> #include <cstdio> #include <cmath> using namespace std; int n ,…