题目描述
At Bessie's recent birthday party, she received N (2 <= N <= 100,000; N%2 == 0) pearls, each painted one of C different colors (1 <= C <= N).
Upon observing that the number of pearls N is always even, her creative juices flowed and she decided to pair the pearls so that each pair of pearls has two different colors.
Knowing that such a set of pairings is always possible for the supplied testcases, help Bessie perform such a pairing. If there are multiple ways of creating a pairing, any solution suffices.
在Bessie最近的生日聚会上，她收到N（2<=N<=100,000; N％2==0）珍珠，每个都涂上C种不同颜色之一（1<=C<=N）。
观察到珍珠N的数量总是均匀的，她的创意来了，决定配对珍珠，使每双珍珠有两种不同的颜色。数据保证存在答案。请帮助Bessie执行这样的配对，如果有多种创建配对的方法，任意输出即可（不过这里没有spj）。
输入格式
* Line 1: Two space-separated integers: N and C
* Lines 2..C + 1: Line i+1 tells the count of pearls with color i: C_i
行1：两个空格分隔的整数：N和C。
行2…C+1：行i+1为颜色i：C_i的珍珠数。
输出格式
* Lines 1..N/2: Line i contains two integers a_i and b_i indicating that Bessie can pair two pearls with respective colors a_i and b_i.
行1…N/2：行i包含两个整数a_i和b_i，表示Bessie可以将两个珍珠与各自的颜色a_i和b_i配对。
输入输出样例
8 3 2 2 4
1 3 1 3 2 3 3 2
说明/提示
There are 8 pearls and 3 different colors. Two pearls have color I; two have color II; four have color III.
Bessie pairs each pearl of color III with one of color I and II.
说明 有8颗珍珠和3种不同的颜色。两个珍珠颜色为1; 两个颜色为2; 四个颜色为3。
Bessie将每种颜色3的珍珠与一种颜色1和2配对。
#include#include #include #include #include #include using namespace std;int s[100001];int n,c,k,x;int main(){ scanf("%d%d",&n,&c); for(int i = 1;i <= c; i++){ scanf("%d",&x); for(int j = 1;j <= x; j++){ k++; s[k] = i; } } sort(s + 1,s + 1 + n); for(int i = 1;i <= n / 2; i++){ printf("%d %d\n",s[i],s[i + (n / 2)]); } return 0;}