poj 3046 Ant Counting

Posted 2020-08-05 Jason杰

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