题意描述
You are given an array a consisting of n positive integers.
Initially, you have an integer x=0. During one move, you can do one of the following two operations:
Choose exactly one i from 1 to n and increase ai by x (ai:=ai x), then increase x by 1 (x:=x 1). Just increase x by 1 (x:=x 1). The first operation can be applied no more than once to each i from 1 to n.
Your task is to find the minimum number of moves required to obtain such an array that each its element is divisible by k (the value k is given).
You have to answer t independent test cases.
给定n个数字，你有一个x，可以对x进行两种操作，询问最少的操作次数，使得n个数都为k的倍数。
思路
我们可以从取模的性质来考虑，如果一个数%k为0，那么不用进行任何操作，如果出现了%k不为0，很明显可以看出需要进行的操作次数为k*余数出现的次数-xi。
AC代码
代码语言：javascript复制#include<bits/stdc .h>
#define x first
#define y second
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long LL;
const int N=2*1e5 10;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
int a[N];
map<LL,LL> cnt;
void solve(){
cnt.clear();
int n,k;cin>>n>>k;
for(int i=0;i<n;i ){
cin>>a[i];
a[i]%=k;
}
LL ans=0;
for(int i=0;i<n;i ){
if(!a[i]) continue;
cnt[a[i]] ;
ans=max(ans,(LL)cnt[a[i]]*k-a[i]);
}
if(ans) ans ;
cout<<ans<<endl;
}
int main(){
IOS;
int t;cin>>t;
while(t--){
solve();
}
return 0;
}
