构造AC的。左右两边都先不用6的倍数，然后哪边数字大那一边往回退一下，然后再比较哪边数字大.......直到结束
#include<cstdio> #include<cstring> #include<cmath> #include<string> #include<vector> #include<queue> #include<algorithm> #include<iostream> using namespace std; const int maxn = 10000000 + 10; int n, m; int a[maxn]; int b[maxn]; int flag[maxn]; int topa, topb; int main() { scanf("%d%d", &n, &m); memset(a, 0, sizeof a); memset(b, 0, sizeof b); memset(flag, 0, sizeof flag); int num1 = 0, num2 = 0, num; int nowa = 0, nowb = 0; for (int aa = 0, i = 2; aa < n; i = i + 2) { if (i % 6 != 0) a[i] = 1, topa = i, aa++; else nowa = i; } for (int aa = 0, i = 3; aa<m; i = i + 3) { if (i % 6 != 0) b[i] = 1, topb = i, aa++; else nowb = i; } while (1) { if (nowa == 0 && nowb == 0) break; if (topa>topb) { a[topa] = 0; topa--; while (1) { if (b[nowa] == 0) { a[nowa] = 1; nowa = nowa - 6; while (1) { if (a[topa]) break; topa--; } break; } nowa = nowa - 6; } } else { b[topb] = 0; topb--; while (1) { if (a[nowb] == 0) { b[nowb] = 1; nowb = nowb - 6; while (1) { if (b[topb]) break; topb--; } break; } nowb = nowb - 6; } } while (1) { if (b[nowa]) nowa = nowa - 6; else break; } while (1) { if (a[nowb]) nowb = nowb - 6; else break; } } printf("%d ", max(topa, topb)); return 0; }