zoukankan      html  css  js  c++  java
  • 3400: [Usaco2009 Mar]Cow Frisbee Team 奶牛沙盘队

    3400: [Usaco2009 Mar]Cow Frisbee Team 奶牛沙盘队

    Time Limit: 3 Sec  Memory Limit: 128 MB
    Submit: 129  Solved: 84
    [Submit][Status][Discuss]

    Description

        农夫顿因开始玩飞盘之后,约翰也打算让奶牛们享受飞盘的乐趣.他要组建一只奶牛飞盘
    队.他的N(1≤N≤2000)只奶牛,每只部有一个飞盘水准指数Ri(1≤Ri≤100000).约翰要选出1只或多于1只奶牛来参加他的飞盘队.由于约翰的幸运数字是F(1≤F≤1000),他希望所有奶牛的飞盘水准指数之和是幸运数字的倍数.
        帮约翰算算一共有多少种组队方式.

    Input

        第1行输入N和F,之后N行输入Ri.

    Output

     
        组队方式数模10^8取余的结果.

    Sample Input

    4 5
    1
    2
    8
    2

    Sample Output

    3

    HINT

        组队方式有(2,3),(3,4),(1,2,4)共三种

    Source

    Silver

    题解:一开始还在想着怎么DFS剪枝= =。。。感觉自己水水哒

    其实就是一个水水的DP就完事啦,记得最后要-1(你总不能一个都不要然后还要算一种方法吧)

     1 /**************************************************************
     2     Problem: 3400
     3     User: HansBug
     4     Language: Pascal
     5     Result: Accepted
     6     Time:260 ms
     7     Memory:23704 kb
     8 ****************************************************************/
     9  
    10 const p=100000000;
    11 var
    12    i,j,k,l,m,n:longint;
    13    b:array[0..5000] of longint;
    14    a:array[0..3000,0..2000] of longint;
    15 begin
    16      readln(n,m);
    17      for i:=1 to n do readln(b[i]);
    18      for i:=1 to n do b[i]:=b[i] mod m;
    19      fillchar(a,sizeof(a),0);
    20      a[0,0]:=1;
    21      for i:=1 to n do
    22          for j:=0 to m do
    23              a[i,j]:=(a[i-1,(j-b[i]+m) mod m]+a[i-1,j]) mod p;
    24      writeln(a[n,0]-1);
    25      readln;
    26 end.   
  • 相关阅读:
    MVC模式-----struts2框架(2)
    MVC模式-----struts2框架
    html的<h>标签
    jsp脚本元素
    LeetCode "Paint House"
    LeetCode "Longest Substring with At Most Two Distinct Characters"
    LeetCode "Graph Valid Tree"
    LeetCode "Shortest Word Distance"
    LeetCode "Verify Preorder Sequence in Binary Search Tree"
    LeetCode "Binary Tree Upside Down"
  • 原文地址:https://www.cnblogs.com/HansBug/p/4418755.html
Copyright © 2011-2022 走看看