来源是《C语言名题精选百则》
算法思想是：
(1)每次从数中抽取一个残留数sum。如果结尾不是1，sum为1，如果结尾是1，sum是所有1的和加1.
(2)抽取sum后原来的数变成size。再用size去分sum。
(3)sum%size作为最后一个数。
/* ------------------------------------------------------ */ /* PROGRAM integer partition : */ /* Given a positive integer n, this program lists all */ /* partition of n in anti-lexical order. */ /* */ /* Copyright Ching-Kuang Shene July/21/1989 */ /* ------------------------------------------------------ */ #include <stdio.h> #include <stdlib.h> /* for atoi() */ #define MAXSIZE 20 void display(int[], int[], int); void main(void) { int partition[MAXSIZE + 1]; /* the actuall partition */ int mult[MAXSIZE + 1]; /* multiplicity */ int part_no; /* no. of parts */ int sum, size, remainder, count; int n; char line[100]; printf("\nPartition of Integer"); printf("\n===================="); printf("\n\nInput a Positive Integer --> "); gets(line); n = atoi(line); partition[1] = n; /* at the biginning, we have*/ mult[1] = part_no = count = 1; /* only one part.*/ display(partition, mult, part_no); do { /* size one sum in 'sum' */ sum = (partition[part_no] == 1) ? mult[part_no--] + 1 : 1; size = partition[part_no] - 1; /* dec. size */ if (mult[part_no] != 1) /* last part with mul=1*/ mult[part_no++]--; /* NO, cut this part */ partition[part_no] = size; /* set new part=size */ mult[part_no] = sum / size + 1; /* fill other*/ if ((remainder = sum % size) != 0) { partition[++part_no] = remainder; mult[part_no] = 1; } count++; display(partition, mult, part_no); } while (mult[part_no] != n); printf("\n\nThere are %d partitions in anti-lexical order", count); system("pause"); } /* ------------------------------------------------------ */ /* FUNCTION display : */ /* This routine displays the given partition. */ /* ------------------------------------------------------ */ void display(int partition[], int mult[], int part_no) { int i, j; printf("\n"); for (i = 1; i <= part_no; i++) /* for each part */ for (j = 1; j <= mult[i]; j++) /* and its mult. */ printf("=", partition[i]); /* show them */ }