http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3216 乱搞的...watashi是分块做的...但我并不知道什么是分块...大概就是把结果相同的数据合并计算 打表跑了一下...发现重复出现的数字很多...于是直接找出会发生重复的数乘起来就行了... /********************* Template ************************/ #include <set> #include <m…

Problem Description For two integers m and k, k is said to be a container of m if k is divisible by m. Given 2 positive integers n and m (m < n), the function f(n, m) is defined to be the number of containers of m which are also no greater than n. Fo…

Number of Containers Time Limit: 1 Second Memory Limit: 32768 KB For two integers m and k, k is said to be a container of m if k is divisible by m. Given 2 positive integers n and m (m < n), the function f(n, m) is defined to be the number of contain…

Number Game Time Limit: 2 Seconds      Memory Limit: 65536 KB The bored Bob is playing a number game. In the beginning, there are n numbers. For each turn, Bob will take out two numbers from the remaining numbers, and get the product of them. There i…

Number Puzzle Time Limit: 2 Seconds      Memory Limit: 65536 KB Given a list of integers (A1, A2, ..., An), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given…

题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2829 题目描述: Mike is very lucky, as he has two beautiful numbers, 3 and 5. But he is so greedy that he wants infinite beautiful numbers. So he declares that any positive number which is…

Number Steps Time Limit: 2 Seconds      Memory Limit: 65536 KB Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (…

一道比较不错的BFS+DP题目 题意很简单,就是问一个刚好包含m(m<=10)个不同数字的n的最小倍数. 很明显如果直接枚举每一位是什么这样的话显然复杂度是没有上限的,所以需要找到一个状态表示方法: 令F[i][j] 表示已经用了 i (二进制压位表示)用了 i 这些数字,且余数j为的状态,枚举时直接枚举当前位,那么答案明显就是F[m][0] 我这里将状态i, j存在了一维空间里,即 i * 1000 + j表示,实际上用一个结构体存队列里的点,用二维数组标记状态也是可行的. #include…

题目 思路: 先倒推!到最后第二步,然后: 初始状态不一定满足这个状态.所以我们要先从初始状态构造出它出发的三种状态.那这三种状态跟倒推得到的状态比较即可. #include<stdio.h> #include<string.h> #include <algorithm> using namespace std; ],b[]; int main() { scanf("%d",&t); while(t--) { scanf(],&a[]…

题面 lcm(x,y)=xy/gcd(x,y) lcm(x1,x2,···,xn)=lcm(lcm(x1,x2,···,xn-1),xn) #include <bits/stdc++.h> using namespace std; long long n,m; ]; long long gcd(long long a,long long b) { if(!b){ return a; } return gcd(b,a%b); } long long lcm(long long a,long lo…