zoukankan      html  css  js  c++  java
  • ZOJ 3662 Math Magic 第37届ACM/ICPC长春赛区H题(DP)

    Math Magic

    Time Limit: 3 Seconds      Memory Limit: 32768 KB

    Yesterday, my teacher taught us about math: +, -, *, /, GCD, LCM... As you know, LCM (Least common multiple) of two positive numbers can be solved easily because of a * b = GCD (a, b) * LCM (a, b).

    In class, I raised a new idea: "how to calculate the LCM of K numbers". It's also an easy problem indeed, which only cost me 1 minute to solve it. I raised my hand and told teacher about my outstanding algorithm. Teacher just smiled and smiled...

    After class, my teacher gave me a new problem and he wanted me solve it in 1 minute, too. If we know three parameters N, M, K, and two equations:

    1. SUM (A1, A2, ..., Ai, Ai+1,..., AK) = N
    2. LCM (A1, A2, ..., Ai, Ai+1,..., AK) = M

    Can you calculate how many kinds of solutions are there for Ai (Ai are all positive numbers). I began to roll cold sweat but teacher just smiled and smiled.

    Can you solve this problem in 1 minute?

    Input

    There are multiple test cases.

    Each test case contains three integers N, M, K. (1 ≤ N, M ≤ 1,000, 1 ≤ K ≤ 100)

    Output

    For each test case, output an integer indicating the number of solution modulo 1,000,000,007(1e9 + 7).

    You can get more details in the sample and hint below.

    Sample Input

    4 2 2
    3 2 2
    

    Sample Output

    1
    2
    

    Hint

    The first test case: the only solution is (2, 2).

    The second test case: the solution are (1, 2) and (2, 1).


    Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

    很水的DP,长春赛的时候竟然没有做出来。。。。。方向是写对了的,只是最后半个小时写的,很紧张。最后超时了。

    需要注意些细节,一些初始化才不会超时。

    预处理出LCM[1000][1000]来。

    dp[now][i][j]表示当前状态下,和为i,LCM为j的解的个数。递推K次就出答案了。

    #include<stdio.h>
    #include<algorithm>
    #include<string.h>
    #include<iostream>
    using namespace std;
    const int MOD=1000000007;
    int dp[2][1010][1010];
    
    int num[1000];
    
    int gcd(int a,int b)
    {
        if(b==0)return a;
        return gcd(b,a%b);
    }
    int lcm(int a,int b)
    {
        return a/gcd(a,b)*b;
    }
    int LCM[1010][1010];
    int main()
    {
        int n,m,k;
        for(int i=1;i<=1000;i++)
          for(int j=1;j<=1000;j++)
            LCM[i][j]=lcm(i,j);
    
    
        while(scanf("%d%d%d",&n,&m,&k)!=EOF)
        {
            int cnt=0;
            for(int i=1;i<=m;i++)
            {
                if(m%i==0)
                   num[cnt++]=i;
            }
            int now=0;
            //memset(dp[now],0,sizeof(dp[now]));
            for(int i=0;i<=n;i++)
              for(int j=0;j<cnt;j++)
                dp[now][i][num[j]]=0;
            dp[now][0][1]=1;
    
            for(int t=1;t<=k;t++)
            {
                now^=1;
               // memset(dp[now],0,sizeof(dp[now]));
              for(int i=0;i<=n;i++)
                for(int j=0;j<cnt;j++)
                  dp[now][i][num[j]]=0;
                for(int i=t-1;i<=n;i++)
                  for(int j=0;j<cnt;j++)
                  {
                      if(dp[now^1][i][num[j]]==0)continue;
                      for(int p=0;p<cnt;p++)
                      {
                          int x=i+num[p];
                          //int y=lcm(num[j],num[p]);
                          int y=LCM[num[j]][num[p]];
                          if(x>n||m%y!=0)continue;
                          dp[now][x][y]+=dp[now^1][i][num[j]];
                          dp[now][x][y]%=MOD;
                      }
                  }
            }
            printf("%d\n",dp[now][n][m]);
        }
        return 0;
    }
    人一我百!人十我万!永不放弃~~~怀着自信的心,去追逐梦想
  • 相关阅读:
    Linux Shell入门
    Linux系统结构
    Ubuntu bond 配置
    VXLAN概述
    lsof
    入驻博客园,希望可以跟大家一起讨论,一起学习和进步。
    版本管理工具小乌龟TortoiseGit的安装和使用(2)
    版本管理工具小乌龟TortoiseGit的安装和使用(1)
    定义变量时未初始化赋值的问题
    BlackBerry 9900刷机
  • 原文地址:https://www.cnblogs.com/kuangbin/p/2727590.html
Copyright © 2011-2022 走看看