POJ 3046 Ant Counting(dp)

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:

3 5 2 3
1
2
2
1
3

Sample Output:

10

题目链接

$dp[i][j]$dp[i][j]代表前i个家族可以组合出元素个数为j的集合个数

递推公式:$dp[i][j]=\sum_{k=0}^{amount[i]}dp[i-1][j-k]$dp[i][j]=∑k=0amount[i]​dp[i−1][j−k]，代表前i-1个家族组合成j-k的集合中每一个集合都加入家族i的蚂蚁k只，相加得到前i个家族组合成j的集合个数。

AC代码:

#pragma comment(linker, "/STACK:102400000,102400000")
//#include <bits/stdc++.h>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <climits>
#include <stdlib.h>
#include <deque>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <string>
#include <vector>
#include <bitset>
#include <complex>
#include <functional>
#include <fstream>
#include <ctime>
#include <stdexcept>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4+5;
const int mod = 1e6;
const double eps = 1e-8;
const double pi = asin(1.0)*2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x){Finish_read=0;x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;Finish_read=1;}

int t, a, s, b;
int x;
int cnt;
int ans;
int dp[maxn][maxn];
map<int, int> amount;

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	read(t); read(a); read(s); read(b);
	for (int i = 1; i <= a; ++i) {
		read(x);
		amount[x]++;
	}
	dp[0][0] = 1;
	cnt = amount[0];	//前i个家族的蚂蚁总数
	for (int i = 1; i <= t; ++i) {
		cnt += amount[i];
		for (int k = 0; k <= amount[i]; ++k) {
			for (int j = cnt; j >= k; --j) {
				dp[i][j] += dp[i - 1][j - k];
				dp[i][j] %= mod;
			}
		}
	}
	ans = 0;
	for (int i = s; i <= b; ++i) {
		ans += dp[t][i];
		ans %= mod;
	}
	printf("%d\n", ans);
#ifndef ONLINE_JUDGE
	fclose(stdin);
	fclose(stdout);
	system("gedit out.txt");
#endif
    return 0;
}

标签：Counting,int,ants,Ant,read,ch,POJ,include,dp
来源： https://blog.csdn.net/Tony5t4rk/article/details/81087092