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

最新推荐文章于 2017-10-19 18:36:00 发布

原创最新推荐文章于 2017-10-19 18:36:00 发布 · 275 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

B. Divisiblity of Differences

Problem Statement

You are given a multiset of n integers. You should select exactly k of them in a such way that the difference between any two of them is divisible by m, or tell that it is impossible.
Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.

Input

First line contains three integers n, k and m (2 ≤ k ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.
Second line contains n integers a1, a2, …, an (0 ≤ ai ≤ $10^9$ ) — the numbers in the multiset.

Output

If it is not possible to select k numbers in the desired way, output «No» (without the quotes).
Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print k integers b1, b2, …, bk — the selected numbers. If there are multiple possible solutions, print any of them.

Examples

Example 1
    Input
        3 2 3
        1 8 4
    Output
        Yes
        1 4
Example 2
    Input
        3 3 3
        1 8 4
    Output
        No
Example 3
    Input
        4 3 5
        2 7 7 7
    Output
        Yes
        2 7 7

题意

给出n个数，再给出k和m，问你能否选出k个数使得这个集合里任意两个数的差都被m整除。

思路

只要对这n个数在%m的意义下统计就行了，任意一个集合多于k个就是Yes..不然就是No。

Code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
inline void readLong(ll &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
/*================Header Template==============*/
multiset<int> divi[100010];
int n,m,k,x;
int main() {
    readInt(n);
    readInt(m);
    readInt(k);
    for(int i=1;i<=n;i++) {
        readInt(x);
        divi[x%k].insert(x);
    }
    int idx=-1;
    for(int i=0;i<k;i++)
        if(divi[i].size()>=m) {
            idx=i;
            break;
        }
    if(idx==-1) {
        puts("No");
        return 0;
    }
    puts("Yes");
    int tot=0;
    for(multiset<int>::iterator it=divi[idx].begin();it!=divi[idx].end()&&tot<m;it++,tot++)
        printf("%d ",*it);
    printf("\n");
    return 0;
}