POJ3046-Ant Counting

最新推荐文章于 2020-02-18 16:01:31 发布

原创最新推荐文章于 2020-02-18 16:01:31 发布 · 357 阅读

0 ·

CC 4.0 BY-SA版权

POJ 同时被 2 个专栏收录

159 篇文章

订阅专栏

----线性dp

66 篇文章

订阅专栏

探讨了给定不同家族蚂蚁数量时，如何计算特定大小群体的所有可能组合数。使用动态规划解决，通过滚动数组优化内存使用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ant Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5746		Accepted: 2157

Description

Bessie was poking around the ant hill one day watching the ants march to and fro while gathering food. She realized that many of the ants were siblings, indistinguishable from one another. She also realized the sometimes only one ant would go for food, sometimes a few, and sometimes all of them. This made for a large number of different sets of ants!

Being a bit mathematical, Bessie started wondering. Bessie noted that the hive has T (1 <= T <= 1,000) families of ants which she labeled 1..T (A ants altogether). Each family had some number Ni (1 <= Ni <= 100) of ants.

How many groups of sizes S, S+1, ..., B (1 <= S <= B <= A) can be formed?

While observing one group, the set of three ant families was seen as {1, 1, 2, 2, 3}, though rarely in that order. The possible sets of marching ants were:

3 sets with 1 ant: {1} {2} {3}
5 sets with 2 ants: {1,1} {1,2} {1,3} {2,2} {2,3}
5 sets with 3 ants: {1,1,2} {1,1,3} {1,2,2} {1,2,3} {2,2,3}
3 sets with 4 ants: {1,2,2,3} {1,1,2,2} {1,1,2,3}
1 set with 5 ants: {1,1,2,2,3}

Your job is to count the number of possible sets of ants given the data above.

Input

* Line 1: 4 space-separated integers: T, A, S, and B

* Lines 2..A+1: Each line contains a single integer that is an ant type present in the hive

Output

* Line 1: The number of sets of size S..B (inclusive) that can be created. A set like {1,2} is the same as the set {2,1} and should not be double-counted. Print only the LAST SIX DIGITS of this number, with no leading zeroes or spaces.

Sample Input

Sample Output

Hint

INPUT DETAILS:

Three types of ants (1..3); 5 ants altogether. How many sets of size 2 or size 3 can be made?

OUTPUT DETAILS:

5 sets of ants with two members; 5 more sets of ants with three members

Source

USACO 2005 November Silver

题意：有a只蚂蚁来自t个家族，选出一些蚂蚁，数量大于等于s小于等于b，问有几种选法

解题思路：先记录每种家族蚂蚁的数量，dp[i][j]表示前i个家族选出j个蚂蚁的方案数，后来发现爆内存，改为了滚动数组

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <cmath>  
#include <map>  
#include <cmath>  
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  

using namespace std;

#define LL long long  
const int INF = 0x3f3f3f3f;

int dp[2][100020];
int x[1020];

int main()
{
	int n, m, s, b;
	while (~scanf("%d%d%d%d", &m, &n, &s, &b))
	{
		memset(x, 0, sizeof x);
		memset(dp, 0, sizeof dp);
		dp[0][0] = 1;
		int sum = 0,a,now=0;
		for (int i = 1; i <= n; i++) scanf("%d", &a), x[a]++;
		for (int i = 1; i <= m; i++)
		{
			sum += x[i];
			memset(dp[now ^ 1], 0, sizeof dp[now ^ 1]);
			for (int j = 0; j <= x[i]; j++)
				for (int k = sum; k >= j; k--)
					dp[now^1][k] = (dp[now][k - j] + dp[now^1][k]) % 1000000;
			now = now ^ 1;
		}
		int ans = 0;
		for (int i = s; i <= b; i++)
			ans = (ans + dp[now][i]) % 1000000;
		printf("%d\n", ans);
	}
	return 0;
}