【Dividing Orange】【CodeForces - 244A】(思維)
題目:
One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 to n·k.
There were k children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the i
Now the children wonder, how to divide the orange so as to meet these conditions:
- each child gets exactly n orange segments;
- the i-th child gets the segment with number a
- no segment goes to two children simultaneously.
Help the children, divide the orange and fulfill the requirements, described above.
Input
The first line contains two integers n, k (1 ≤ n, k ≤ 30). The second line contains kspace-separated integers a
It is guaranteed that all numbers ai are distinct.
Output
Print exactly n·k distinct integers. The first n integers represent the indexes of the segments the first child will get, the second n integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.
You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.
Examples
Input
2 2 4 1
Output
2 4 1 3
Input
3 1 2
Output
3 2 1
解題報告:存在一個橘子,可以分成n*k瓣,有k個孩子,每個人都有了自己一定要有的瓣數(標記)的,問每個的分配情況。就是在一定的區間裡面,成功的將所有的瓣數分配出去,利用mark標記就可以實現。
ac程式碼:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1005;
int num[35];
int flag[maxn];
int main()
{
int n,k;
scanf("%d%d",&n,&k);
int l=0;
memset(num,0,sizeof(num));
memset(flag,0,sizeof(flag));
for(int i=0;i<k;i++)
{
scanf("%d",&num[i]);
flag[num[i]-1]=1;
}
for(int i=0;i<k;i++)
{
printf("%d ",num[i]);
for(int j=0;j<n-1;)
{
if(!flag[l])
{
printf("%d ",1+l);
j++;
flag[l]=1;
}
l++;
}
printf("\n");
}
return 0;
}