给定N是一个正整数，求比N大的最小“不重复数”，这里的不重复是指没有两个相等的相邻位，如1102中的11是相等的两个相邻位故不是不重复数，而12301是不重复数。

/* *给定N是一个正整数，求比N大的最小“不重复数”，这里的不重复是指没有两个相等的相邻位，如1102中的11是相 *等的两个相邻位故不是不重复数，而12301是不重复数。 */#include <stdio.h>unsigned int solve(unsigned int k); unsigned int testify(unsigned int k);intmain(void){unsigned int num;while (1) {scanf("%d", &num);if (num == 0)break;printf("->");solve(num);}/*testify*/for ( num = 1; num < 111111; ++num) {if (solve(num) != testify(num)) {printf("num = %u, solve = %u, testify = %u\n", num, solve(num), testify(num));break;}}return 0;}intis_num(unsigned int k) {if (k < 11)return 1;int first = k % 10;int second;k = k / 10;while (k) {second = k % 10;if (second == first)return 0;first = second;k = k / 10;}return 1;}unsigned intsolve(unsigned int k){char str[100];int  start, end, mark;if (k == 0 || k == (unsigned int)-1)return;++k;if (is_num(k)) {printf("%d\n", k);return k;}sprintf(str + 1, "%d", k);start = 1;end   = 2;while (str[end] != '\0') {if (str[end] == str[end -1]) {if (str[end] == '9') {str[end - 1] = str[end] = '0';if (end  == 2) {start = 0;str[0] = '1';mark = 1;break;} str[end - 2] = str[end - 2] + 1;end -= 2;} else {str[end] = str[end] + 1;mark = end +1;break;}} else {++end;}}/*返回整数是为了下面的验证*/k = 0;for( ; start != mark; ++start) {printf("%c", str[start]);k = 10 * k + str[start] - '0';}while (str[start++] != '\0') {printf("%c", '0');k = 10 * k;if (str[start++] != '\0') {printf("%c", '1');k = 10 * k + 1;}elsebreak;}printf("\n");return k;}unsigned inttestify(unsigned int k){while (!is_num(++k));return k;}