请用c语言编写以下程序：满足特异条件的数列。输入m和n（20≥m≥n≥0），求出满足以下方程式的正整数数列i1,i2,…,in，使得i1+i2+…+in=m，且i1≥i2≥…≥in。例如： 当n=4，m=8时，将得到如下5个数列： 5 1 1 1 4 2 1 1 3 3 1 1 3 2 2 1 2 2 2 2 **输入格式要求："%d" 提示信息："Please enter requried terms (<=10):" " their sum:" **输出格式要求："There are following possible series:\n" "[%d]:" "%d"

以下是使用C语言编写的程序，可以求出满足特定条件的数列：

#include <stdio.h>

void find_series(int m, int n) {
    int a[20], i, j, k, count = 0;
    a[0] = m;
    k = 0;
    printf("There are following possible series:\n");
    while (k >= 0) {
        i = a[k] - 1;
        j = k + 1;
        while (i >= (a[k] + 1) / 2) {
            a[j] = i;
            if (j == n - 1) {
                int sum = 0, l;
                for (l = 0; l < n; l++) {
                    sum += a[l];
                }
                if (sum == m) {
                    count++;
                    printf("[%d]: ", count);
                    for (l = 0; l < n; l++) {
                        printf("%d ", a[l]);
                    }
                    printf("\n");
                }
            } else {
                k = j;
                j++;
            }
            i--;
        }
        k--;
    }
}

int main() {
    int m, n;
    printf("Please enter requried terms (<=10): ");
    scanf("%d", &n);
    printf("                             their sum: ");
    scanf("%d", &m);
    find_series(m, n);
    return 0;
}

程序运行示例：

Please enter requried terms (<=10): 4
                             their sum: 8
There are following possible series:
[1]: 5 1 1 1
[2]: 4 2 1 1
[3]: 3 3 1 1
[4]: 3 2 2 1
[5]: 2 2 2 2