贪心 HDOJ 5355 Cake

好的，数据加强了，wa了

题目传送门

 /*
     题意：1到n分成m组，每组和相等
     贪心：先判断明显不符合的情况，否则肯定有解(可能数据弱？)。贪心的思路是按照当前的最大值来取
         如果最大值大于所需要的数字，那么去找没有vis的数字
 */
 /************************************************
  * Author        :Running_Time
  * Created Time  :2015-8-6 13:24:55
  * File Name     :C.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 bool vis[MAXN];
 vector<int> ans[];
 ll n, m;

 int main(void)    {     //HDOJ 5355 Cake
     int T;  scanf ("%d", &T);
     while (T--) {
         memset (vis, false, sizeof (vis));
         scanf ("%I64d%I64d", &n, &m);
         ll tot = n * (n + ) / ;
         if (tot % m !=  || tot / m < n)    {
             puts ("NO");    continue;
         }

         ll sum = tot / m;   ll mx = n;
         for (int i=; i<=m; ++i)    {
             ans[i].clear ();    ll tmp = sum;
             while (tmp >= mx)   {
                 tmp -= mx;  ans[i].push_back (mx);  vis[mx] = true;
                 while (mx -  >=  && vis[mx])  --mx;
             }
             ll t = mx;
             while (tmp != )    {
                 while ((t -  >=  && vis[t]) || tmp < t)  t--;
                 tmp -= t;   ans[i].push_back (t);   vis[t] = true;
                 t = tmp;
             }
         }

         puts ("YES");
         for (int i=; i<=m; ++i)    {
             printf ("%d", (int) ans[i].size ());
             for (int j=; j<ans[i].size (); ++j)    {
                 printf (" %d", ans[i][j]);
             }
             puts ("");
         }
     }

     return ;
 }