poj 3046 Ant Counting

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

#include <cstdio>
#include <cstdlib>
#include <cstring>
#define M 1000000
int cnt[1002] = {0};
int c1[100002] = {0};
int c2[100002] = {0};
int ans[100002] = {0};
int n;

void multi(int cn1, int cn2) {
    for(int i = 0; i <= cn1; i++) {
        for(int j = 0; j <= cn2; j++) {
            ans[i+j] = (ans[i+j] + c1[i]*c2[j])%M;
        }
    }
    n = cn1 + cn2;
}

int main(int argc, char const *argv[])
{
    //freopen("input.txt","r",stdin);
    int t, s, a, b;
    scanf("%d %d %d %d",&t, &a, &s, &b);
    for(int i = 0; i < a; i++) {
        int tmp;
        scanf("%d",&tmp);
        cnt[tmp]++;
    }
    for(int i = 0; i <= cnt[1];i++) {
        ans[i] = 1;
    }
    n = cnt[1];
    for(int i = 2; i <= t; i++) {
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        for(int j = 0; j <= n; j++) {
            c1[j] = ans[j];
        }
        for(int j = 0; j <= cnt[i]; j++) {
            c2[j] = 1;
        }
        memset(ans, 0, sizeof(ans));
        multi(n,cnt[i]);
    }
    int ansm = 0;
    for(int i = s; i <= b; i++) {
        ansm = (ansm + ans[i])%M;
    }
    printf("%d
",ansm);
    return 0;
}

注意，此题要求输出后六位，要mod M 

此题也可以用动态规划法求解

dp[i][j]表示前i个家族构成j的方法数

那么

dp[i][j] = dp[i-1][j-k](k..0 to min(j,c[i]));

上面的式子可写为
dp[i][j] = dp[i][j-1] + dp[i-1][j]-dp[i][j-1-c[i]];

代码如下

/*dp[i][j] means i to j geshu 

dp[i][j] = dp[i-1][j-k](k..0 to min(j,c[i]));
dp[i][j] = dp[i][j-1] + dp[i-1][j]-dp[i][j-1-c[i]];
    
}*/


#include <cstdio>
#include <cstdlib>
#include <cstring>
#define M 1000000
int cnt[1002] = {0};
int dp[1002][100002] = {0};


int main(int argc, char const *argv[])
{
    freopen("input.txt","r",stdin);
    int t, s, a, b;
    scanf("%d %d %d %d",&t, &a, &s, &b);
    for(int i = 0; i < a; i++) {
        int tmp;
        scanf("%d",&tmp);
        cnt[tmp]++;
    }

    for(int i = 0; i <= t; i++) {
        dp[i][0] = 1;
    }

    for(int i = 0; i < t; i++) {
        for(int j = 1; j <= b; j++) {
            if(j-1-cnt[i+1] >= 0) {
                dp[i+1][j] = (dp[i+1][j-1] + dp[i][j]-dp[i][j-1-cnt[i+1]] + M ) % M;
            }
            else {
                dp[i+1][j]  = (dp[i+1][j-1] + dp[i][j])%M;
            }
        }
    }
    
    int ansm = 0;
    for(int i = s; i <= b; i++) {
        ansm = (ansm + dp[t][i] + M)%M;
    }
    printf("%d
",ansm);
    return 0;
}