Codeforces 509 B Painting Pebbles 贪心

读题目是一个坎。。。真心的。。。看了大半天才看题目的意思。。。

传送门：http://codeforces.com/contest/509/problem/B

题意：

讲的是你有n堆石头 有k种颜色可以涂 第二行是代表每堆石头中石头的数量

题目要求你给所有石头涂色。要求一堆石头和另外一堆石头中相同颜色的石头的数量之差小于等于1 

举第一个例子

1

1 4

1 2 4

1 2 3 4

第一堆石头中有颜色1 的一个 其他都为0个

第四堆石头中有颜色1 2 3 4的石头各一个 

这样颜色1 的石头的数量差  为0  其他的为1 

符合题意。所以是YES

B. Painting Pebbles

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Sample test(s)

input

4 41 2 3 4

output

YES11 41 2 41 2 3 4

input

5 23 2 4 1 3

output

NO

input

5 43 2 4 3 5

output

YES1 2 31 31 2 3 41 3 41 1 2 3 4

解法：贪心。。想要达到目标。。就需要尽可能把颜色平摊。这样才能得到最优解。

判断一堆堆石头中石头数最少的和最多的之间的差  如果差大于你有的颜色的数量 那样肯定没有解。输出NO  

其他输出YES  然后按照颜色依次输出就好   比如   1 2 3 4 ....  n 1 2 3 4 ...n  

AC代码

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<cmath>#include<queue>#define INF 0x0f0f0f0fusing namespace std;int main(){int i,j,k,l,m,n,imax=0,imin=105;int a[105];scanf("%d%d",&n,&m);for(i=0;i<n;i++){scanf("%d",&a[i]);if(a[i]>imax)imax=a[i];if(a[i]<imin)imin=a[i];}if(imax-imin>m){printf("NO\n");}else{printf("YES\n");for(i=0;i<n;i++){for(j=0;j<a[i]-1;j++){printf("%d ",j%m+1);}printf("%d ",j%m+1);printf("\n");}}}