数论整除——cf1059D

用map是卡着过去的。。题解用vector+离散化后常数小了十倍。。

总之就是把所有模数给保存下来然后离散化，再去匹配一下即可，最后有个细节

自己的

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define maxn 200005
ll n,k,a[maxn];
map<ll,ll>mp[11];
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        ll tmp=a[i],len=0;
        while(tmp)len++,tmp/=10;
        mp[len][a[i]%k]++; 
    }
    ll ans=0;
    for(int i=1;i<=n;i++){
        ll tmp=a[i],tmp2=a[i],len=0;
        while(tmp2)len++,tmp2/=10;
        for(int j=1;j<=10;j++){
            tmp=tmp*10%k;
            ans+=mp[j][(k-tmp)%k];
            if(j==len && a[i]%k==(k-tmp)%k)ans--;
        }
        
    }
    cout<<ans<<endl;
}

题解的

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

typedef long long li;

using namespace std;

const int N = 200 * 1000 + 13;
const int LOGN = 11;

int n, k;
int a[N];
int len[N];
vector<int> rems[LOGN];
int pw[LOGN];

int main() {
    scanf("%d%d", &n, &k);
    forn(i, n) scanf("%d", &a[i]);
    
    pw[0] = 1;
    forn(i, LOGN - 1)
        pw[i + 1] = pw[i] * 10 % k;
    
    forn(i, n){
        int x = a[i];
        while (x > 0){
            ++len[i];
            x /= 10;
        }
        rems[len[i]].push_back(a[i] % k);
    }
    
    forn(i, LOGN)
        sort(rems[i].begin(), rems[i].end());
    
    li ans = 0;
    forn(i, n){
        for (int j = 1; j < LOGN; ++j){
            int rem = (a[i] * li(pw[j])) % k;
            int xrem = (k - rem) % k;
            auto l = lower_bound(rems[j].begin(), rems[j].end(), xrem);
            auto r = upper_bound(rems[j].begin(), rems[j].end(), xrem);
            ans += (r - l);
            if (len[i] == j && (rem + a[i] % k) % k == 0)
                --ans;
        }
    }
    
    printf("%lld\n", ans);
    return 0;
}

标签：forn,数论,rems,int,cf1059D,ans,LOGN,整除,pw
来源： https://www.cnblogs.com/zsben991126/p/10811605.html