zoukankan      html  css  js  c++  java
  • B. Divisiblity of Differences

    B. Divisiblity of Differences
    time limit per test1 second
    memory limit per test512 megabytes
    inputstandard input
    outputstandard output
    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 ≤ 109) — 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.

    如果集合中两两之差能被m整除,那么它们%m之后的余数应该相等。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<queue>
     4 #include<algorithm>
     5 #include<cmath>
     6 #include<ctime>
     7 #include<set>
     8 #include<map>
     9 #include<stack>
    10 #include<cstring>
    11 #define inf 2147483647
    12 #define For(i,a,b) for(register int i=a;i<=b;i++)
    13 #define p(a) putchar(a)
    14 #define g() getchar()
    15 //by war
    16 //2017.11.1
    17 using namespace std;
    18 int n,k,m;
    19 int a[100010];
    20 int b[100010];
    21 int l;
    22 int cnt;
    23 
    24 void in(int &x)
    25 {
    26     int y=1;
    27     char c=g();x=0;
    28     while(c<'0'||c>'9')
    29     {
    30     if(c=='-')
    31     y=-1;
    32     c=g();
    33     }
    34     while(c<='9'&&c>='0')x=(x<<1)+(x<<3)+c-'0',c=g();
    35     x*=y;
    36 }
    37 void o(int x)
    38 {
    39     if(x<0)
    40     {
    41         p('-');
    42         x=-x;
    43     }
    44     if(x>9)o(x/10);
    45     p(x%10+'0');
    46 }
    47 int main()
    48 {
    49     in(n),in(k),in(m);
    50     For(i,1,n)
    51     {
    52     in(a[i]);    
    53     b[a[i]%m]++;
    54     if(cnt<b[a[i]%m])
    55     {
    56     cnt=b[a[i]%m];    
    57     l=a[i]%m;
    58     }
    59     }
    60     if(cnt<k)
    61     {
    62         puts("No");
    63         exit(0);
    64     }
    65     puts("Yes");
    66     For(i,1,n)
    67     {
    68         if(a[i]%m==l)
    69         {
    70             k--;
    71             o(a[i]),p(' ');
    72         }
    73         if(k==0)
    74         break;
    75     }
    76      return 0;
    77 }
  • 相关阅读:
    U1
    std::vector的内存释放
    (转)浅谈PostgreSQL的索引
    PostgreSQL的generate_series函数应用
    linux 里的`反引号
    wireshark抓取本地回环数据包
    linux shell中 if else以及大于、小于、等于逻辑表达式介绍
    c++利用互斥锁实现读写锁
    c++互斥锁的实现
    大小端,"字节序"
  • 原文地址:https://www.cnblogs.com/war1111/p/7768530.html
Copyright © 2011-2022 走看看