UVA - 110 Meta-Loopless Sorts

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题目大意：模仿 Pascal 输出 n 个数字比较大小的过程。

解题思路：全排列，向上层序列从后往前插入新元素，如给出 n=3，则最开始为 a，第一层 b，插入为 ab，到第二层 c，为 abc，向前得到 acb，cab，然后返回上一层 b，向前移动为 ba，此基础上第二层bac，bca，cba，返回上一层 b，结束，回到 a，结束。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>using namespace std;char ans[10];int n;void dfs(int now) {    if (now == n) {        for (int i = 0; i < now; i++) printf("  ");        printf("writeln(");        for (int i = 0; i < n-1; i++)            printf("%c,", ans[i]);        printf("%c)\n", ans[n-1]);        return;    }    int used[10];    for (int i = 0; i < now; i++)        used[i] = ans[i];    for (int i = now+1; i > 0; i--) {        ans[i] = ans[i-1];        ans[i-1] = 'a'+now;        if (now > 0) {            for (int j = 0; j < now; j++) printf("  ");            if (i == now+1) printf("if %c < %c then\n", ans[i-2], now+'a');            else if (i == 1) printf("else\n");            else printf("else if %c < %c then\n", ans[i-2], ans[i-1]);        }        dfs(now+1);    }    for (int i = 0; i < now; i++)        ans[i] = used[i];}int main() {    int T;    scanf("%d", &T);    while (T--) {        scanf("%d", &n);        printf("program sort(input,output);\nvar\n");        for (int i = 0; i < n; i++)            ans[i] = 'a' + i;        for (int i = 0; i < n-1; i++)            printf("%c,", ans[i]);        printf("%c : integer;\nbegin\n  readln(", ans[n-1]);        for (int i = 0; i < n-1; i++)            printf("%c,", ans[i]);        printf("%c);\n", ans[n-1]);        dfs(0);        printf("end.\n");        if (T) printf("\n");    }return 0;}

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