int main() { int x; while (~scanf ("%d", &x)) { if ((x & 1) || x == 2) puts ("NO"); else puts ("YES"); } return 0; }

int main() { int n, sum, i, a[35], b[35], x, y; while (~scanf ("%d%d", &n, &sum)) { x = y = 0; memset (a, 0, sizeof(a)); for (i = 0; i < n; i++) { scanf ("%d%d", a+i, b+i); x += a[i], y += b[i]; //x为最小和，y为最大和，可能形成的和为[x, y] } for (i = n - 2; i >= 0; i--) a[i] += a[i+1]; //a[i] = (a[i] + a[i+1] + … + a[n-1]) if (sum >= x && sum <= y) { puts ("YES"); for (i = 0; i < n; i++) { if (i > 0) printf (" "); int tp = min (b[i], sum-a[i+1]); sum -= tp; printf ("%d", tp); } printf ("\n"); } else puts ("NO"); } return 0; }

C题
题意：输入一个用户名进行注册，如果该用户名之前木有被注册过，输出"OK"，如果注册过在后面加个编号i，i从1开始，还存在，那么i++，直到可以注册为止，输出这个可以注册的用户名
注意一下这组数据：

用map暴力搞搞即可

map<string, int> m; char ch[100005], s[100005+M]; //必须开够哦 int main() { int n, k, i, j, tp; while (~scanf ("%d", &n)) { m.clear(); while (n--) { scanf ("%s", s); if (m[s] == 0) puts ("OK"); else { tp = m[s]++; //tp获取s之前出现的次数，然后出现次数+1 k = 0; while (tp) { ch[k++] = tp % 10 + '0'; tp /= 10; } ch[k] = 0; //tp转为字符串加到s的后面 j = strlen (s); for (i = k - 1; i >= 0; i--) s[j++] = ch[i]; s[j] = 0; printf ("%s\n", s); } m[s]++; //s的出现次数+1 } } return 0; }

struct Dp{ int v; int next; //记录路径 }dp[M]; struct envelope{ int w, h, i; }e[M]; bool cmp (envelope a, envelope b) { if (a.w == b.w) return a.h > b.h; return a.w > b.w; } int main() { int n, w, h, i, j, maxs, v, num; while (~scanf ("%d%d%d", &n, &w, &h)) { num = 1; for (i = 0; i < n; i++) { scanf ("%d%d", &e[i].w, &e[i].h); e[i].i = num++; //编号num独立，不受下面的if影响 if (e[i].w <= w || e[i].h <= h) i--, n--; } if (n <= 0) //没有信件可以装下卡片 { puts ("0"); continue; } for (i = 0; i < n; i++) //初始化 dp[i].v = 1, dp[i].next = -1; maxs = 1; sort (e, e+n, cmp); v = 0; for (i = 0; i < n; i++) //单调最长递减子序列模型，逆向方便记录路径 { for (j = i + 1; j < n; j++) { if (e[i].h > e[j].h && e[i].w > e[j].w && dp[j].v < dp[i].v + 1) { dp[j].v = dp[i].v + 1; dp[j].next = i; if (maxs < dp[j].v) maxs = dp[j].v, v = j; } } } printf ("%d\n", maxs); printf ("%d", e[v].i); for (i = dp[v].next; i != -1; i = dp[i].next) printf (" %d", e[i].i); printf ("\n"); } return 0; }

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