B. Modulo Sum time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a sequence of numbers a1, a2, ..., an, and a number m. Check if it is possible to choose a non-empty subsequence…

题目传送:http://acm.hdu.edu.cn/showproblem.php?pid=5776 Problem Description Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO Input The first line of the input has a…

sum 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5776 Description Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO Input The first line of the input has an i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5776 题目让你求是否有区间的和是m的倍数. 预处理前缀和,一旦有两个数模m的值相同,说明中间一部分连续子列可以组成m的倍数. //#pragma comment(linker, "/STACK:102400000, 102400000") #include <algorithm> #include <iostream> #include <cstdlib>…

题意:给定 n 个数,和 m,问你是不是存在连续的数和是m的倍数. 析:考虑前缀和,如果有两个前缀和取模m相等,那么就是相等的,一定要注意,如果取模为0,就是真的,不要忘记了,我当时就没记得.... 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #i…

分析:就是判断简单的前缀有没有相同,注意下自身是m的倍数,以及vis[0]=true; #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vect…

猜了一下,发现对了.n>m是一定有解的.所以最多m*m暴力,一定能找到.而T较小,所以能过. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #…

Sum Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/131072K (Java/Other) Total Submission(s) : 78   Accepted Submission(s) : 30 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Sample Input 2 Sample Output 2…

sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1126    Accepted Submission(s): 494 Problem Description Given a sequence, you're asked whether there exists a consecutive subsequence whose…

链接:传送门 题意:给一个长为 n 的串,问是否有子串的和是 m 的倍数. 思路:典型鸽巢定理的应用,但是这里 n,m 的大小关系是不确定的,如果 n >= m 根据定理可以很简单的判定是一定有解的,当 n < m 的时候就需要去具体寻找一下了,这里构造一个新串 Si = a1 + a2 + a3 + ...... + ai ,如果新串 Si % m = 0 自然就yes了,对于任意一个串 Si % m 的余数范围在 [ 0 , m - 1 ] ,如果出现两个余数相同的新串 S 则就能构成 (…