题目链接:http://lightoj.com/volume_showproblem.php?problem=1197 题意:给你两个数 a b,求区间 [a, b]内素数的个数, a and b (1 ≤ a ≤ b < 231, b - a ≤ 100000). 由于a和b较大,我们可以筛选所有[2, √b)内的素数,然后同时去筛选掉在区间[a, b)的数,用IsPrime[i-a] = 1表示i是素数: ///LightOj1197求区间素数的个数; #include<stdio.h&g…
题目链接:传送门 题目: Prime Distance Time Limit: 1000MS Memory Limit: 65536K Total Submissions: Accepted: Description The branch of mathematics called number theory and itself). The first prime numbers are ,,, but they quickly become less frequent. One of the…
头文件:#include <math.h> fmod() 用来对浮点数进行取模(求余),其原型为: double fmod (double x); 设返回值为 ret,那么 x = n * y + ret,其中 n 是整数,ret 和 x 有相同的符号,而且 ret 的绝对值小于 y 的绝对值.如果 x = 0,那么 ret = NaN. fmod 函数计算 x 除以 y 的 f 浮点余数,这样 x = i*y + f,其中 i 是整数,f 和 x 有相同的符号,而且 f 的绝对值小于…
这个相对于两个大整数的运算来说,只能说是,low爆了. 只要利用好除法的性质,这类题便迎刃而解.O(∩_∩)O哈哈~ //大整数除一个int数 #include<iostream> #include<cstdio> #include<cstring> using namespace std; char s[1000],result[1000]; int main() { long long divis; int n,i,k,flag,len; char c; while…
Problem Description Everybody knows any number can be combined by the prime number. Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. Specially, LPF(1) = 0. In…
题目大意: 给出T个实例,T<=200,给出[a,b]区间,问这个区间里面有多少个素数?(1 ≤ a ≤ b < 231, b - a ≤ 100000) 解题思路: 由于a,b的取值范围比较大,无法把这个区间内的所以素数全部筛选出来,但是b-a这个区间比较小,所以可以用区间素数筛选的办法解决这个题目. 代码: #include<cstdio> #include<cstring> #include<iostream> #include<algorith…
D. Soldier and Number Game time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the seco…
