贪心 Codeforces Round #289 (Div. 2, ACM ICPC Rules) B. Painting Pebbles

题目传送门

 /*
     题意：有 n 个piles，第 i 个 piles有 ai 个pebbles，用 k 种颜色去填充所有存在的pebbles，
             使得任意两个piles，用颜色c填充的pebbles数量之差 <= 1。
             如果不填充某种颜色，就默认数量为0。
     1. 贪心：如果个数之间超过k个，那么填充什么颜色都会大于1，巧妙地思维
         详细解释：http://blog.csdn.net/haoliang94/article/details/43672617
     2. 比较每种填充颜色在n组里使用最多和最少的，若差值<=1，则YES，否则NO
         详细解释：http://www.cnblogs.com/windysai/p/4265469.html
 */
 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <map>
 using namespace std;

 const int MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];

 int main(void)
 {
     #ifndef ONLINE_JUDGE
     freopen ("B.in", "r", stdin);
     #endif

     int n, k;
     while (~scanf ("%d%d", &n, &k))
     {
         int mx = -;    int mn = INF;
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i]);
             if (mx < a[i])    mx = a[i];
             if (mn > a[i])    mn = a[i];
         }

         if (mx - mn <= k)
         {
             puts ("YES");
             for (int i=; i<=n; ++i)
             {
                 int t = ;
                 for (int j=; j<=a[i]; ++j)
                 {
                     printf ("%d%c", ++t, (j==a[i]) ? '\n' : ' ');
                     if (t == k)    t = ;
                 }
             }
         }
         else
             puts ("NO");
     }

     return ;
 }
 

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <map>
 #include <vector>
 using namespace std;

 const int MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];
 int cnt[MAXN];
 vector <int> v[MAXN];

 int main(void)
 {
     #ifndef ONLINE_JUDGE
         freopen ("B.in", "r", stdin);
     #endif

     int n, k;
     while (~scanf ("%d%d", &n, &k))
     {
         for (int i=; i<=MAXN; ++i)
             v[i].clear ();

         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i]);
             memset (cnt, , sizeof (cnt));
             int t = ;
             for (int j=; j<=a[i]; ++j)
             {
                 cnt[++t]++;
                 if (t == k)    t = ;
             }
             for (int j=; j<=k; ++j)
             {
                 v[j].push_back (cnt[j]);
             }
         }

         bool flag = true;
         vector<int>:: iterator it1, it2;
         for (int i=; i<=k; ++i)
         {
             sort (v[i].begin (), v[i].end ());
             it1 = v[i].end () - ;
             it2 = v[i].begin ();
             if (*it1 - *it2 > )
             {
                 flag = false;    break;
             }
         }

         if (flag)
         {
             puts ("YES");
             for (int i=; i<=n; ++i)
             {
                 int t = ;
                 for (int j=; j<=a[i]; ++j)
                 {
                     printf ("%d%c", ++t, (j==a[i]) ? '\n' : ' ');
                     if (t == k)    t = ;
                 }
             }
         }
         else
             puts ("NO");
     }

     return ;
 }

解法2