zoukankan      html  css  js  c++  java
  • HDU 4248 A Famous Stone Collector 组合数学dp ****

    A Famous Stone Collector

    Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 793    Accepted Submission(s): 292


    Problem Description
    Mr. B loves to play with colorful stones. There are n colors of stones in his collection. Two stones with the same color are indistinguishable. Mr. B would like to
    select some stones and arrange them in line to form a beautiful pattern. After several arrangements he finds it very hard for him to enumerate all the patterns. So he asks you to write a program to count the number of different possible patterns. Two patterns are considered different, if and only if they have different number of stones or have different colors on at least one position.
     
    Input
    Each test case starts with a line containing an integer n indicating the kinds of stones Mr. B have. Following this is a line containing n integers - the number of
    available stones of each color respectively. All the input numbers will be nonnegative and no more than 100.
     
    Output
    For each test case, display a single line containing the case number and the number of different patterns Mr. B can make with these stones, modulo 1,000,000,007,
    which is a prime number.
     
    Sample Input
    3
    1 1 1
    2
    1 2
     
    Sample Output
    Case 1: 15
    Case 2: 8
    Hint
    In the first case, suppose the colors of the stones Mr. B has are B, G and M, the different patterns Mr. B can form are: B; G; M; BG; BM; GM; GB; MB; MG; BGM; BMG; GBM; GMB; MBG; MGB.
     
    Source
     
    Recommend
     
    题意:给n种石头,每种m个。
            求组成长度 小于等于k的全排列的个数。
    解题思路:
     
    别人的思路:

      dp[ i ][ j ]表示:考虑前i种石头构成的长度为j的序列的个数。
      转台转移方程:
        dp[ i ][ j ] = dp[ i-1 ][ j ];   //未放入第i种颜色的石头
        for  k := 1 ~ min( j , s[ i ] )  //放入k个第i种颜色的石头
          dp[ i ][ j ] += dp[ i-1 ][ j - k ] * C[ j ][ k ]; //!!!
          Cnm = Cn-1m-1 + Cn-1 1
       其中C[ n ][ m ]表示组合数。
     
     
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cstdlib>
     5 using namespace std;
     6 typedef __int64 LL;
     7 const LL mod=1000000007;
     8 
     9 LL a[102];
    10 LL cnm[10002][102];
    11 LL  dp[102][10002];
    12 void Init()
    13 {
    14     LL i,j;
    15     for(i=0;i<=10000;i++)
    16         cnm[i][0]=1;
    17     for(i=1;i<=10000;i++)
    18     {
    19         for(j=1;j<=i&&j<=100;j++)
    20         {
    21             if(i==j) cnm[i][j]=1;
    22             else cnm[i][j]=( cnm[i-1][j]+cnm[i-1][j-1] )%mod;
    23         }
    24     }
    25 }
    26 void solve(LL n)
    27 {
    28     LL i,j,s,Sum=0,k;
    29     memset(dp,0,sizeof(dp));
    30     for(i=0;i<=100;i++)
    31         dp[i][0]=1;
    32     for(i=1,s=0;i<=n;i++)
    33     {
    34         s=s+a[i];
    35         for(j=1;j<=s;j++)
    36         {
    37             dp[i][j]=dp[i-1][j];
    38             for(k=1;k<=a[i];k++)
    39                 if(k<=j)
    40                 dp[i][j]=( dp[i][j]+(cnm[j][k]*dp[i-1][j-k])%mod )%mod;
    41             if(i==n) Sum=(Sum+dp[i][j])%mod;
    42         }
    43     }
    44     printf("%I64d
    ",Sum);
    45 }
    46 int main()
    47 {
    48     LL T=0;
    49     LL i,n;
    50     Init();
    51     while(scanf("%I64d",&n)>0)
    52     {
    53         for(i=1;i<=n;i++)
    54             scanf("%I64d",&a[i]);
    55         printf("Case %I64d: ",++T);
    56         solve(n);
    57     }
    58     return 0;
    59 }
  • 相关阅读:
    Python笔记_第四篇_高阶编程_再议装饰器和再议内置函数
    Python笔记_第四篇_高阶编程_实例化方法、静态方法、类方法和属性方法概念的解析。
    Python笔记_第四篇_高阶编程_二次封装
    Python笔记_第四篇_高阶编程_反射_(getattr,setattr,deattr,hasattr)
    Python笔记_第四篇_高阶编程_正则表达式_3.正则表达式深入
    Python笔记_第四篇_高阶编程_正则表达式_2.正则表达式入门
    Python笔记_第四篇_高阶编程_正则表达式_1.正则表达式简介(re模块)
    Python笔记_第四篇_高阶编程_检测_2.文档检测
    愿你的眼中总有光芒,活成你想要的模样!
    ruby-rails 环境搭建
  • 原文地址:https://www.cnblogs.com/tom987690183/p/3454101.html
Copyright © 2011-2022 走看看