poj1012约瑟夫问题
推出递推公式 f = (f + m - 1) % (2 * k - i);
若超时，需打表；
# include <stdio.h># include <stdlib.h># include <string># include <string.h>int s[33];int out(int m, int k) {int point = -1;int t = m;int tmp = k;while (k --) {m = t;while(m --) {while (!s[(++ point) % (s[30])]);}point = point % (s[30]);//if (s[point] <= tmp) return 0;printf ("%d ", s[point]);s[point] = 0;}return 1;}int pre (int k) {for (int i = 0; i < k * 2; ++ i) {s[i] = i + 1;}}int go (int n, int &point, int k) {while (n --) {while (!s[(++ point) % (k * 2)]);}if(point >= 2 * k)point -= 2 * k;int t = s[point];s[point] = 0;return t;}int check(int m, int k) {int tmp = 0;for (int i = k * 2; i > k; -- i) {tmp = (tmp + m - 1) % i;if (tmp < k) return 0;}return 1;}int main () {int k;memset(s, 0, sizeof(int) * 30);for(;;) {scanf ("%d", &k);if (!k) return 0;//s[30] = k * 2;int flag = 0;int i;for (int x = 1; ; x++) {if (check(x * (k + 1),k)) {printf("%d\n", x * (k + 1)); break;};if (check(x*(k+1) + 1,k)) {printf("%d\n", x* (k + 1) + 1); break;}/*flag = 0;i = x * (k + 1);here://int point = -1;int j;int tmp = 0;//if (tmp <= k && tmp > 0) continue;//pre(k);//tmp = 0;for (j = 0; j < k; ++ j) {tmp = (i + tmp - 1) % (k * 2 - j);//if (tmp == 0) tmp = 2 * k - j;if (tmp < k)break;//if (go(tmp, point, k) <= k) break;}if (j == k) {printf ("%d ", i);//pre(k);//out(i, k);break;}if (!flag) {i = x * (k + 1) + 1; flag = 1;goto here; }*//*int t = i % (k * 2);if (t <= k && t > 0) continue;//printf ("%d ", i);go (k);if ( out(i, k) ) {printf ("%d\n", i);break;}*/}}return 0;}