Krypton Factor UVA

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每次取一个元素，符合就进行下次的选取就成。

唯一需要注意的是本题的格式必须看仔细，题目中讲的很详细：As such a sequence is potentially very long, split it into groups of four (4) characters separated bya space. If there are more than 16 such groups, please start a new line for the 17th group.别搞错啦

#include<cstdio>#include<cstring>int a[88], b[88], ser, len;void set_in(int l, int cnt) {if(ser == 0) {len = cnt;return;}for(int i=0; i<cnt; i++) a[i] = b[i];for(int i=0; i<l; i++) {int cur = 1;bool ok = true;a[cnt] = i;while(cur <= (cnt+1)/2) {ok = false;for(int j=0; j<cur; j++) if(a[cnt-j]!=a[cnt-j-cur]) ok=true;if(!ok) break;cur++;}if(ok) {if(ser) ser--; elsereturn;b[cnt] = i;set_in(l, cnt+1);}}}int main() {int l;const int zu = 16 * 4;while(scanf("%d%d", &ser, &l) && ser) {memset(a, 0, sizeof(a));set_in(l, 0);for(int i=0; i<len; i++) {printf("%c", b[i]+'A');if(i == len-1) {printf("\n");continue;}if(i%zu==(zu-1)) {printf("\n");continue;}if(i%4==3) printf(" ");}printf("%d\n", len);}}

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