杭电1019Least Common Multiple

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杭电1019Least Common Multiple

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1019 题目: Problem Description The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 an…

杭电1019-Least Common Multiple

#include<stdio.h>int gcd(int a,int b);int main(){ int n,m,a,b,i,sum;//sum是最小公倍数 scanf("%d",&n); while(n--) { scanf("%d",&m); sum=1;//sum=1 for(i=1;i<=m;i++) …

HDU——1019Least Common Multiple（多个数的最小公倍数）

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42735 Accepted Submission(s): 16055 Problem Description The least common multiple (LCM) of a set of positive integers is…

HDU - 1019-Least Common Multiple(求最小公倍数（gcd）)

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105. Input Input will consist of multiple problem instances. The f…

杭电 1159 Common Subsequence

Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a stri…

杭电1159 Common Subsequence【最长公共子序列】

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 解题思路:任意先给出两个字符串 abcfbc abfcab,用dp[i][j]来记录当前最长的子序列,则如果有x[i]与y[j]相等的话,则相当于公共子序列的长度在dp[i-1][j-1]上增加1, 如果x[i]与y[j]不相等的话,那么dp[i][j]就取得dp[i][j-1]和dp[i-1][j]中的最大值即可.时间复杂度为O(mn) 反思:大概思路想出来之后,因为dp数组赋初值调了很久,…

（杭电1019 最小公倍数） Least Common Multiple

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 64855 Accepted Submission(s): 24737 Problem Description The least common multiple (LCM) of a set of positive integers is the sm…