POJ 3046 Ant Counting

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

 

 

背包dp

可以优化决策，只贴了优化后的代码。

 

#include<cstdio>

int cnt[1005]={};
int dp[2][100005];
const int MOD = 1000000;

int main(){
    int T, A, S, B, x;
    scanf("%d%d%d%d",&T,&A,&S,&B);
    for(int i=0;i<A;i++)
        scanf("%d",&x),cnt[x]++;

    // dp[i][j] 前 i 种中选 j 个物品可以组成的种数
    // 决策： 不选i 或 至少选一个i
    // 不选的话 显然就是 dp[i-1][j]
    // 选至少一个 dp[i][j-1] - dp[i-1][j-1-cnt[i]]
    dp[0][0] = 1;
    for(int i=1;i<=T;i++){
        dp[i&1][0] = 1;
        for(int j=1;j<=B;j++){
            if(j-cnt[i] > 0)// 无法用单独使用第 i 件物品构建出 j
                dp[i&1][j] =
                (dp[(i+1)&1][j]+dp[i&1][j-1]-dp[(i+1)&1][j-1-cnt[i]]+MOD)%MOD;
            else
                dp[i&1][j] = (dp[(i+1)&1][j] + dp[i&1][j-1])%MOD;
            // 第 i 件物品 >= j 所以可以用 j-1 件 i 之前的物品加上一件 i 构建 j
            // 可以单独用 j-1 件物品（这些物品可以由 k 组成 k <= i）
            // 加上一件 i 构建 j

        }
    }
    int ans = 0;
    for(int i=S;i<=B;i++)
        ans += dp[T&1][i];
    printf("%d\n",ans%MOD);
    return 0;
}