Codeforces Round #441 (Div. 2, by Moscow Team Olympiad) B. Divisiblity of Differences

http://codeforces.com/contest/876/problem/B

题意：

给出n个数，要求从里面选出k个数使得这k个数中任意两个的差能够被m整除，若不能则输出no。

思路：

差能够被m整除，其实就是对m取余的余数相同。那么就统计n个数的余数丢到一个map里面，最后判断是否有某个数的数量大于等于k。

代码：

#include <stdio.h>
#include <map>
#include <vector>
using namespace std;

map<int,int> mmp;

int a[100005];
int b[100005];

vector<int> v;

int main()
{
    int n,k,m;

    scanf("%d%d%d",&n,&k,&m);

    for (int i = 0;i < n;i++)
    {
        int t;

        scanf("%d",&t);

        b[i] = t;

        t %= m;

        a[i] = t;
    }

    for (int i = 0;i < n;i++) mmp[a[i]]++;

    map<int,int>::iterator it;

    int ans = -1;

    for (it = mmp.begin();it != mmp.end();++it)
    {
        if (it -> second >= k)
        {
            ans = it -> first;
            break;
        }
    }

    if (ans == -1) printf("No
");
    else
    {
        printf("Yes
");

        int cnt = 0;

        for (int i = 0;i < n;i++)
        {
            if (b[i] % m == ans)
            {
                v.push_back(b[i]);
                cnt++;
            }

            if (cnt >= k) break;
        }

        for (int i = 0;i < v.size();i++)
        {
            printf("%d%c",v[i],i == v.size() - 1 ? '
' :' ');
        }
    }

    return 0;
}