• SPOJ AMR10I Dividing Stones


    Time limit: 7s Source limit: 50000B Memory limit: 256MB





     
    The first line contains the number of test cases T. T lines follow, one corresponding to each test case, containing 2 integers: N and P. 
     
    OUTPUT  


     
    EXPLANATION
    In the first test case, the possible ways of division are (1,1,1), (1,2), (2,1) and (3) which have values 1, 2, 2, 3 and hence, there are 3 distinct values. 
    In the second test case, the numbers 1 to 6 constitute the answer and they can be obtained in the following ways: 
    1=1*1*1*1*1 
    2=2*1*1*1 
    3=3*1*1 
    4=4*1 
    5=5 
    6=2*3
    题意:有n个石子,可以分成任意堆,每一种分法的值为每一堆的石子数量的乘积。求一共可以分成多少个不同的乘积。
    分析:最终的乘积除了1以外,都可以分解成素数相乘或者素数相乘再与1相乘的形式。因为n不超过70,所以我们可以先找出不超过70的所有素数,然后从这些素数中进行搜索求解即可。为了方便求出不同的乘积有多少个,可以用STL里面的set来统计不同的数有多少个。
     
     
     
    #include<cstdio>
    #include<cstring>
    #include<set>
    #include<algorithm>
    using namespace std;
    typedef long long LL;
    set<LL> s;
    int prime[21] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73};
    int n, p;
    
    void dfs(int num, int cur, LL ans)
    {
        s.insert(ans);
        if(cur < prime[num]) return ;
        dfs(num, cur - prime[num], ans * prime[num] % p); //要第num个素数
        dfs(num+1, cur, ans); //不要第num个素数
    }
    
    int main()
    {
        int T, i, j;
        scanf("%d",&T);
        while(T--) {
            scanf("%d%d",&n,&p);
            s.clear();
            dfs(0, n, 1);
            printf("%d
    ", s.size());
        }
        printf("
    ");
    }
  • 相关阅读:
    Cocos 更新时反复杀进程,导致差异更新失效的Bug
    Cocos 编译android-studio
    Could not resolve all dependencies for configuration ':classpath'
    MAC Gradle 下载的问题
    命令行创建 keystore
    应用间共享文件 FileProvider
    Android jks 签名文件 生成
    Gradle 离线 安装
    信息
    Cocos Lua的Touch 点击事件添加
  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4690275.html
Copyright © 2020-2023  润新知