题目链接

http://noi.openjudge.cn/ch0206/9289/

描述

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.

输入

* 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

输出

* 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.

样例输入

3 5 2 3
1
2
2
1
3

样例输出

10

 　　英文题目也是真的难懂，理解了很久，其实题目意思挺简单的，就是给你s个t种蚂蚁，每种蚂蚁有一定的数量，然后去求用这些种蚂蚁组成s——b之间长的，情况数有多少。

　　仔细理解的话可以发现，我们可以把这个看成背包问题，对于这t种蚂蚁，如果其中一组数量为k，那么就有k+1种情况，既不选这种蚂蚁，选一个，选两个……选k个。

　　那么我们就套用背包模板，先枚举种类，再枚举序列长度，第三层枚举选的个数。每种区间，第三层枚举的选不同个数的情况之和，那么状态转移方程就是

　　　　f[i][j]+=f[i-1][j-k]   //f[i][j]表示前i个种类中选j长的序列的情况数。

所以当第i种选k个时，那么情况数就是前i-1种选j-k个的情况数，因为然后再枚举k，k去不同数时的情况加起来，就是前i种选j长的序列的情况数。

注意：一定要给f数组整上一个开始的值，不然怎么加都是0；所以我们把前i种选0的情况赋值成1。

最后附上代码

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int n,a,s,b,num[1010],f[1010][100010];
 5 int main()
 6 {
 7     scanf("%d%d%d%d",&n,&a,&s,&b);
 8     for(int i=1,t;i<=a;++i)
 9     {
10         scanf("%d",&t);
11         num[t]++;
12     }
13     for(int i=0;i<=n;++i)
14     {
15         f[i][0]=1;
16     }
17     for(int i=1;i<=n;++i)
18     {
19         for(int j=1;j<=b;++j)
20         {
21             for(int k=0;k<=num[i];++k)
22             {
23                 if(j>=k)
24                 f[i][j]=f[i][j]+f[i-1][j-k];
25             }
26         }
27     }
28     int ans=0;
29     for(int i=s;i<=b;++i)
30     {
31         ans+=f[n][i];
32     }
33     printf("%d",ans);
34     return 0;
35 }