POJ 3046 Ant Counting（dp----多重集组合数）

题目链接：http://poj.org/problem?id=3046

Ant Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4763		Accepted: 1816

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

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


题意：有n中蚂蚁，第i种蚂蚁有N_i个，一共有A个蚂蚁。不同类别的蚂蚁可以相互区分，但同种类别的蚂蚁不能相互区别。从这些蚂蚁中分别取出S，S+1...B个，一共有多少种取法。(解释不清楚，请看题中的Hint)

两个优化，dp[i+1][j]=sigma(dp[i][j-k])=dp[i+1][j-1]+dp[i][j]-dp[i][j-1-Mi]

另外一个就是开滚动数组减小空间复杂度

具体可以看《挑战程序设计竞赛》P68~69

<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[1010];
int dp[1010][100010];
int t,n,s,b;
int mod=1e6;
int main()
{
    while(~scanf("%d %d %d %d",&t,&n,&s,&b))
    {
        int x;
        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++){
            scanf("%d",&x);
            a[x-1]++;
        }
        for(int i=0;i<=t;i++)//一个都不取的方法只有一种
            dp[i][0]=1;
        for(int i=0;i<t;i++){
            for(int j=1;j<=b;j++){
                if(j-1-a[i]>=0){
                    //在取余的情况下，要避免减法运算的结果出现负数
                    dp[i+1][j]=(dp[i+1][j-1]+dp[i][j]-dp[i][j-1-a[i]]+mod)%mod;
                }
                else
                    dp[i+1][j]=(dp[i+1][j-1]+dp[i][j])%mod;
            }
        }
        int ans=0;
        for(int i=s;i<=b;i++)
            ans=(ans+dp[t][i])%mod;
        printf("%d\n",ans);
    }
    return 0;
}
</span>