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poj 3046 Ant Counting

2024-08-11 06:42:37   211人阅读

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 312213

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 

poj 3046 Ant Counting


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