LINK:HDU 1018 题意:求n!的位数~ 由于n!最后得到的数是十进制,故对于一个十进制数,求其位数可以对该数取其10的对数,最后再加1~ 易知:n!=n*(n-1)*(n-2)*......*3*2*1 ∴lg(n!)=lg(n)+lg(n-1)+lg(n-2)+......+lg(3)+lg(2)+lg(1); 代码: #include <iostream> #include <cstdio> #include <cmath> using namespace…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1018 解题报告:输入一个n,求n!有多少位. 首先任意一个数 x 的位数 = (int)log10(x) + 1; 所以n!的位数 = (int)log10(1*2*3*.......n) + 1; = (int)(log10(1) + log10(2) + log10(3) + ........ log10(n)) + 1; #include<cstdio> #include<cstrin…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35382    Accepted Submission(s): 16888 Problem Description In many applications very large integers numbers are required. Some of these…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21106    Accepted Submission(s): 9498 Problem Description In many applications very large integers numbers are required. Some of these…

题意: 给一个数n,返回该数的阶乘结果是一个多少位(十进制位)的整数. 思路: 用对数log来实现. 举个例子 一个三位数n 满足102 <= n < 103: 那么它的位数w 满足 w = lg103 = 3. 因此只要求lgn 向下取整 +1就是位数.然后因为阶乘比如5阶乘的话是5 * 4 * 3 * 2 * 1.位数就满足lg 5 * 4 * 3 * 2 * 1 = lg5 + lg4 + lg3 + lg2 + lg1.用加法就不会超过数字上限. 当然这是十进制下得.如果是m进制下 ,…

数学题  用的这个方法比较烂 g++超时  c++ 406ms /************************************************************************* > Author: xlc2845 > Mail: xlc2845@gmail.com > Created Time: 2013年10月16日 星期三 20时47分16秒 *****************************************************…

Problem Description Inmany applications very large integers numbers are required. Some of theseapplications are using keys for secure transmission of data, encryption, etc.In this problem you are given a number, you have to determine the number ofdig…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38886    Accepted Submission(s): 18889 Problem Description In many applications very large integers numbers are required. Some of these…

题意是求 n 的阶乘的位数. 直接求 n 的阶乘再求其位数是不行的,开始时思路很扯淡,想直接用一个数组存每个数阶乘的位数,用变量 tmp 去存 n 与 n - 1 的阶乘的最高位的数的乘积,那么 n 的阶乘的位数就等于 n - 1 的阶乘的位数加 tmp 的位数再减去 1. 但这种做法是不对的,例如有可能最高位与 n 的乘积结果是 99,而其实 n 与其他位的乘积结果是能进到这一位的,也就是说实际应该在 n - 1 的阶乘位数上增加 2 ( 3 -1 ) 位.而在对样例测试时也发现 n 为 10…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34084    Accepted Submission(s): 16111 Problem Description In many applications very large integers numbers are required. Some of these…

1. 利用数学公式lg(n!)=lg(2)+lg(3)+....+lg(n) 求解 2.…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40262    Accepted Submission(s): 19637 Problem Description In many applications very large integers numbers are required. Some of these…

最近一堆题目要补,一直咸鱼,补了一堆水题都没必要写题解.备忘一下这个公式. Stirling公式的意义在于:当n足够大时,n!计算起来十分困难,虽然有很多关于n!的等式,但并不能很好地对阶乘结果进行估计,尤其是n很大之后,误差将会非常大.但利用Stirling公式可以将阶乘转化成幂函数,使得阶乘的结果得以更好的估计.而且n越大,估计得越准确. 传送门:_(:з」∠)_ 再来一个详细一点的,传送门:( ･´ω`･ ) Big Number Time Limit: 2000/1000 MS (Jav…

Problem Description In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of…

题目大意: 将一个数开阶乘后得到的值,来求这个值的位数 n! = 1*2*3*4...*n 对于求一个数的位数的方法为ans = lg(n!) + 1 那么就可以看作 ans = lg(1) + lg(2) .......+ lg(n) + 1 #include <cstdio> #include <cmath> ; int main() { int t; scanf("%d" , &t); while(t--){ int n; scanf("…

Problem Description In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of…

Big Number 题意:算n!的位数. 题解:对于一个数来算位数我们一般都是用while去进行计算,但是n!这个数太大了,我们做不到先算出来在去用while算位数. while(a){ cnt++; a/=; } 将一个数对取10对数(取整),然后再加一就是这个数的位数,然后我们在算n!的时候每次对10取对数就好了. #include<iostream> #include<cmath> using namespace std; int main() { ios::sync_wi…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…

斯特灵公式 Wiki http://zh.wikipedia.org/wiki/斯特林公式 /** \brief hdu 1018 * * \param date 2014/7/24 * \param state AC * \return * */ #include <iostream> #include <fstream> #include <cmath> using namespace std; const double PI=3.1415926; int main…

大意就是求 : log10(n!) = log10(1 * 2 *  3 * .......*n) = log10(1) + log10(2) + ........+log10(n); 打表的话会MLE,直接递推即可了.后台数据不会非常刁钻. #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

Leftmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12229    Accepted Submission(s): 4674题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1060 Problem Description Given a positive integ…

hdu 5720 问题描述 黎明时,Venus为Psyche定下了第二个任务.她要渡过河,收集对岸绵羊身上的金羊毛. 那些绵羊狂野不驯,所以Psyche一直往地上丢树枝来把它们吓走.地上现在有n n n根树枝,第i i i根树枝的长度是ai a_i a​i​​. 如果她丢的下一根树枝可以和某两根树枝形成三角形,绵羊就会被激怒而袭击她. 现在Psyche手中只有长度不小于L L L且不大于R R R的树枝.请你帮忙计算,她下一根可以丢多少种不同长度的树枝而不会把绵羊激怒呢? 输入描述 第一行,一个…

HDU   5694 Problem Description 众所周知,度度熊喜欢的字符只有两个:B和D.今天,它发明了一种用B和D组成字符串的规则:S(1)=BS(2)=BBDS(3)=BBDBBDD…S(n)=S(n−1)+B+reverse(flip(S(n−1))其中,reverse(s)指将字符串翻转,比如reverse(BBD)=DBB,flip(s)指将字符串中的B替换为D,D替换为B,比如flip(BBD)=DDB.虽然度度熊平常只用它的电脑玩连连看,这丝毫不妨碍这台机器无与伦比…

http://acm.hdu.edu.cn/showproblem.php?pid=5768 Lucky7 Problem Description   When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its bod…

HDU 5877 Weak Pair(弱点对) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Description 题目描述 You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned. An ordere…

HDU 5833 Zhu and 772002 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math pro…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5446 Unknown Treasure 问题描述 On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere…

Dylans loves numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5272 Description Dylans是谁?你可以在 UOJ 和 Codeforces上看到他.在BestCoder里,他有另外一个ID:s1451900.今天的题目都和他有关哦.Dylans得到了一个数N.他想知道N的二进制中有几组1.如果两个1之间有若干个(至少一个)0…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5720 题意:有n(n <= 105)个数 ,每个数小于等于 1018:问在给定的[L,R]区间中,有多少个数不能与已知的n个数中的任何两个组成三角形? 思路:由三角形的边长关系移位即可得到符合关系的长度len:a[i] - a[j] +1 <= len <= a[i] + a[j] - 1;(a[i] > a[j]) 易知当i固定时,j为i - 1时len符合的区间是最大的: 这就变成…