随笔分类 -  数论

摘要:1148: Prime CutsResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K1346447StandardA prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program that will cut some number of prime numbers from the list of prime numbe 阅读全文
posted @ 2012-05-06 20:24 漂木 阅读(226) 评论(0) 推荐(0)
摘要:1196: GoldbachResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K2275745StandardIn 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:Every even number greater than 4 can be written as the sum of two odd prime num 阅读全文
posted @ 2012-05-06 19:11 漂木 阅读(397) 评论(0) 推荐(0)
摘要:2573: Product of two primesResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s65536K929287Standard输出所有在10000 – 20000之间,恰好可以分解为两个素数乘积的整数。将这些数从小到大输出。Input无Output每一行输出一个符合要求的数,按照从小到大排列。下面是输出的头和尾,中间部分需自行计算10001.........19999Problem Source:skywindThis problem is used for contest:131Submit/Problem List/S 阅读全文
posted @ 2012-05-06 18:37 漂木 阅读(276) 评论(0) 推荐(0)
摘要:1013: Polynomial MultiplicationResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K4660807StandardA polynomial is a sum of terms, where each term is a coefficient multiplied by the variable x raised to some nonnegative integer power. The largest such power is referred to as the degree of the po 阅读全文
posted @ 2012-04-17 21:51 漂木 阅读(387) 评论(0) 推荐(0)
摘要:Georg Cantor用这张数表证明了有理数是可枚举的刘汝佳教主的白书上有这道题,不过貌似解答是错误的T_T1/1 1/2 1/3 1/4 1/5 …2/1 2/2 2/3 2/4 …3/1 3/2 3/3 …4/1 4/2 …5/1 ……网上说Z字型编号。。。我怎么就不觉得是Z字型。。。明明是蛇形1-1/1; 2-1/2; 3-2/1; 4-3/1; 5-2/2; 6-1/3.....如下图输出第N项的值:样例输入:314712345样例输出:2/12/41/459/99 1 #include <stdio.h> 2 3 int main(void) 4 { 5 int n; 阅读全文
posted @ 2012-02-15 21:40 漂木 阅读(348) 评论(0) 推荐(0)