暴力超时版：

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long LL;
const int N = 1e8 + 10;
LL f[N];
int ck(int x) { int res = 0; while (x > 0) res += x % 10, x /= 10; return res; }
int main()
{
    f[0] = 0;
    for (int i = 1; i <= N - 10; i++)
    {
        if (i % 10 == 0) f[i] = ck(i);
        else f[i] = f[i - 1] + 1;
    }
    for (int i = 1; i <= N - 10; i++) f[i] += f[i - 1];
    int T; scanf("%d", &T);
    while (T--)
    {
        int x; scanf("%d", &x);
        printf("%llu\n", f[x]);
    }
    return 0;
}

标程：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e8 + 10;
LL d[9];
int main()
{
    d[0] = 0; d[1] = 45; for (int i = 2, p = 10; i <= 8; i++, p *= 10) d[i] = d[i - 1] * 10 + 45 * p;
    int T; scanf("%d", &T);
    while (T--)
    {
        int n; scanf("%d", &n);
        LL ans = 0;
        int p = 1, n1 = n, n2 = 0, t = 0;
        while (n1 > 0)
        {
            int x = n1 % 10; n1 /= 10;
            
            ans += x * (n2 + 1);
            ans += x * d[t] + (x - 1) * x / 2 * p;
            
            n2 = x * p + n2;

            p = p * 10;
            t++;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

暴力超时版：

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long LL;
const int N=1e8+10;
LL f[N];
int ck(int x){int res=0;while(x>0)res+=x%10,x/=10;return res;}
int main()
{
    f[0]=0;
    for(int i=1;i<=N-10;i++)
    {
        if(i%10==0)f[i]=ck(i);
        else f[i]=f[i-1]+1;
    }
    for(int i=1;i<=N-10;i++)f[i]+=f[i-1];
    int T;scanf("%d",&T);
    while(T--)
    {
        int x;scanf("%d",&x);
        printf("%llu\n",f[x]);
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e8+10;
LL d[9];
int main()
{
    d[0]=0;d[1]=45;for(int i=2,p=10;i<=8;i++,p*=10)d[i]=d[i-1]*10+45*p;
    int T;scanf("%d",&T);
    while(T--)
    {
        int n;scanf("%d",&n);
        LL ans=0;
        int p=1,n1=n,n2=0,t=0;
        while(n1>0)
        {
            int x=n1%10;n1/=10;
        ans+=x*(n2+1);
        ans+=x*d[t]+(x-1)*x/2*p;
        
        n2=x*p+n2;

        p=p*10;
        t++;
    }
    printf("%lld\n"&#44;ans);
}
return 0;

}