题意:给出一列数a,给出m个区间,再给出每个区间的最小公倍数 还原这列数 因为数组中的每个数至少都为1,而且一定是这个区间的最小公约数ans[i]的倍数,求出它与ans[i]的最小公倍数,如果大于1e9(题目中给的范围,一定不能够还原) 最后按照这样算出每一个a[i]后,再检查一遍这m个区间的算出来的最小公约数是否和给出的一致 学习的dzy4939414644的代码 #include<iostream> #include<cstdio> #include<cstring>…

Describtion In mathematics, the greatest common divisor (gcd) of two or more integers, when at least one of them is not zero, is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4.-Wikiped…

GCD Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2742    Accepted Submission(s): 980 Problem Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There ar…

GCD is Funny 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5902 Description Alex has invented a new game for fun. There are n integers at a board and he performs the following moves repeatedly: He chooses three numbers a, b and c written at the boa…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4272    Accepted Submission(s): 1492 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

GCD 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=1695 Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be…

GCD and LCM Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4497 Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4675 题意:给出n,m,K,一个长度为n的数列A(1<=A[i]<=m).对于d(1<=d<=m),有多少个长度为n的数列B满足: (1)1<=B[i]<=m; (2)Gcd(B[1],B[2],……,B[n])=d: (3)恰有K个位置满足A[i]!=B[i]. 思路: i64 p[N]; void init(){    p[0]=1;    int i;    FOR1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5726 给你n个数,q个询问,每个询问问你有多少对l r的gcd(a[l] , ... , a[r]) 等于的gcd(a[l'] ,..., a[r']). 先用RMQ预处理gcd,dp[i][j] 表示从i开始2^j个数的gcd. 然后用map存取某个gcd所对应的l r的数量. 我们可以在询问前进行预处理,先枚举i,以i为左端点的gcd(a[i],..., a[r])的种类数不会超过log2(n)…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5726 [题目大意] 给出数列An,对于询问的区间[L,R],求出区间内数的GCD值,并且求出GCD值与其相等的区间总数 [题解] 首先,固定一个区间的右端点,利用GCD的递减性质,可以求出GCD相等的区间左端点的范围,将其范围的左右端点保存下来,同时,对于每个新产生的区间,以其GCD值为下标的MAP值+1,最后对于每个询问,在其右端点保存的范围中查找,获得其GCD值,同时在MAP中获取该GCD值…