P2946 [USACO09MAR]牛飞盘队Cow Frisbee Team

飞盘团队组合算法

最新推荐文章于 2025-09-11 19:30:36 发布

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最新推荐文章于 2025-09-11 19:30:36 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zzyh/p/7202854.html

针对农夫约翰组建飞盘队伍的问题，此算法采用动态规划求解。给定N只奶牛及其飞盘技能指数，需选出若干只奶牛使得其技能指数之和为约翰幸运数字F的倍数。

题目描述

After Farmer Don took up Frisbee, Farmer John wanted to join in the fun. He wants to form a Frisbee team from his N cows (1 <= N <= 2,000) conveniently numbered 1..N. The cows have been practicing flipping the discs around, and each cow i has a rating R_i (1 <= R_i <= 100,000) denoting her skill playing Frisbee. FJ can form a team by choosing one or more of his cows.

However, because FJ needs to be very selective when forming Frisbee teams, he has added an additional constraint. Since his favorite number is F (1 <= F <= 1,000), he will only accept a team if the sum of the ratings of each cow in the team is exactly divisible by F.

Help FJ find out how many different teams he can choose. Since this number can be very large, output the answer modulo 100,000,000.

Note: about 50% of the test data will have N <= 19.

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．

输入输出格式

输入格式：

Line 1: Two space-separated integers: N and F
Lines 2..N+1: Line i+1 contains a single integer: R_i

输出格式：

Line 1: A single integer representing the number of teams FJ can choose, modulo 100,000,000.

输入输出样例

输入样例#1：

输出样例#1：

说明

FJ has four cows whose ratings are 1, 2, 8, and 2. He will only accept a team whose rating sum is a multiple of 5.

FJ can pair the 8 and either of the 2's (8 + 2 = 10), or he can use both 2's and the 1 (2 + 2 + 1 = 5).

【题目大意】

从n个数中选出几个数使他成为f的倍数的方案数。

【思路】

动态规划...

dp[i][j]表示前i个牛，选择的牛的数的和%f==j的方案数。不是很懂转移方程+r[i]不是-r[i]

【code】

#include<iostream>
#include<cstdio>
using namespace std;
#define mod 100000000
int n,f;
int r[2001],dp[2001][1009];
int main()
{
    scanf("%d%d",&n,&f);
    for(int i=1;i<=n;i++)
    scanf("%d",&r[i]);
    dp[0][0]=1;
    for(int i=1;i<=n;i++)
    for(int j=0;j<=f;j++)
    {
        dp[i][j]+=dp[i-1][j]+dp[i-1][(j+r[i])%f];
        dp[i][j]%=mod;
    }
    printf("%d\n",dp[n][f]%mod);
    return 0;
}