题目大意：
有编号一到t的蚂蚁家族，每个家族有不同的蚂蚁数。
问构成S只蚂蚁到构成B只蚂蚁共有多少种方式
例如
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}
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 这道题用母函数来求解。所谓母函数，就是生成函数，说白了就是多项相乘，每一项分别表示每个家族里取一只，两只，三只。。。最后乘出的结果中，x的n次方的系数就表示构成n个蚂蚁有多少种方式代码如下
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #define M 1000000
5 int cnt[1002] = {0};
6 int c1[100002] = {0};
7 int c2[100002] = {0};
8 int ans[100002] = {0};
9 int n;
10
11 void multi(int cn1, int cn2) {
12 for(int i = 0; i <= cn1; i++) {
13 for(int j = 0; j <= cn2; j++) {
14 ans[i+j] = (ans[i+j] + c1[i]*c2[j])%M;
15 }
16 }
17 n = cn1 + cn2;
18 }
19
20 int main(int argc, char const *argv[])
21 {
22 //freopen("input.txt","r",stdin);
23 int t, s, a, b;
24 scanf("%d %d %d %d",&t, &a, &s, &b);
25 for(int i = 0; i < a; i++) {
26 int tmp;
27 scanf("%d",&tmp);
28 cnt[tmp]++;
29 }
30 for(int i = 0; i <= cnt[1];i++) {
31 ans[i] = 1;
32 }
33 n = cnt[1];
34 for(int i = 2; i <= t; i++) {
35 memset(c1, 0, sizeof(c1));
36 memset(c2, 0, sizeof(c2));
37 for(int j = 0; j <= n; j++) {
38 c1[j] = ans[j];
39 }
40 for(int j = 0; j <= cnt[i]; j++) {
41 c2[j] = 1;
42 }
43 memset(ans, 0, sizeof(ans));
44 multi(n,cnt[i]);
45 }
46 int ansm = 0;
47 for(int i = s; i <= b; i++) {
48 ansm = (ansm + ans[i])%M;
49 }
50 printf("%d\n",ansm);
51 return 0;
52 }
注意，此题要求输出后六位，要mod M