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 = 200 + 10;
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 = -1;    int mn = INF;
        for (int i=1; 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=1; i<=n; ++i)
            {
                int t = 0;
                for (int j=1; j<=a[i]; ++j)
                {
                    printf ("%d%c", ++t, (j==a[i]) ? '
' : ' ');
                    if (t == k)    t = 0;
                }
            }
        }
        else
            puts ("NO");
    }

    return 0;
}

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

const int MAXN = 100 + 10;
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=1; i<=MAXN; ++i)
            v[i].clear ();
        
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &a[i]);
            memset (cnt, 0, sizeof (cnt));
            int t = 0;
            for (int j=1; j<=a[i]; ++j)
            {
                cnt[++t]++;
                if (t == k)    t = 0;
            }
            for (int j=1; j<=k; ++j)
            {
                v[j].push_back (cnt[j]);
            }
        }

        bool flag = true;
        vector<int>:: iterator it1, it2;
        for (int i=1; i<=k; ++i)
        {
            sort (v[i].begin (), v[i].end ());
            it1 = v[i].end () - 1;
            it2 = v[i].begin ();
            if (*it1 - *it2 > 1)
            {
                flag = false;    break;
            }
        }

        if (flag)
        {
            puts ("YES");
            for (int i=1; i<=n; ++i)
            {
                int t = 0;
                for (int j=1; j<=a[i]; ++j)
                {
                    printf ("%d%c", ++t, (j==a[i]) ? '
' : ' ');
                    if (t == k)    t = 0;
                }
            }
        }
        else
            puts ("NO");
    }

    return 0;
}