一眼。
将 \(a\) 中每个数用前导零补到 \(6\) 位，题目相当于问所有 \(i,j\)，\(a_i\) 的每一位加 \(a_j\) 的这一位都不超过 \(9\) 的 \((i,j)\) 对数。
直接高维前缀和统计即可，时间复杂度 \(O(n + 10^6)\)。
code
/*
p_b_p_b txdy
AThousandSuns txdy
Wu_Ren txdy
Appleblue17 txdy
*/
#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 1000100;
int n, a[maxn], b[maxn][9], sum[10][10][10][10][10][10];
void solve() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
int x = a[i];
for (int j = 0; j < 6; ++j) {
b[i][j] = x % 10;
x /= 10;
}
++sum[b[i][0]][b[i][1]][b[i][2]][b[i][3]][b[i][4]][b[i][5]];
}
for (int i = 0; i <= 9; ++i) {
for (int j = 0; j <= 9; ++j) {
for (int k = 0; k <= 9; ++k) {
for (int l = 0; l <= 9; ++l) {
for (int p = 0; p <= 9; ++p) {
for (int q = 1; q <= 9; ++q) {
sum[i][j][k][l][p][q] += sum[i][j][k][l][p][q - 1];
}
}
}
}
}
}
for (int i = 0; i <= 9; ++i) {
for (int j = 0; j <= 9; ++j) {
for (int k = 0; k <= 9; ++k) {
for (int l = 0; l <= 9; ++l) {
for (int p = 1; p <= 9; ++p) {
for (int q = 0; q <= 9; ++q) {
sum[i][j][k][l][p][q] += sum[i][j][k][l][p - 1][q];
}
}
}
}
}
}
for (int i = 0; i <= 9; ++i) {
for (int j = 0; j <= 9; ++j) {
for (int k = 0; k <= 9; ++k) {
for (int l = 1; l <= 9; ++l) {
for (int p = 0; p <= 9; ++p) {
for (int q = 0; q <= 9; ++q) {
sum[i][j][k][l][p][q] += sum[i][j][k][l - 1][p][q];
}
}
}
}
}
}
for (int i = 0; i <= 9; ++i) {
for (int j = 0; j <= 9; ++j) {
for (int k = 1; k <= 9; ++k) {
for (int l = 0; l <= 9; ++l) {
for (int p = 0; p <= 9; ++p) {
for (int q = 0; q <= 9; ++q) {
sum[i][j][k][l][p][q] += sum[i][j][k - 1][l][p][q];
}
}
}
}
}
}
for (int i = 0; i <= 9; ++i) {
for (int j = 1; j <= 9; ++j) {
for (int k = 0; k <= 9; ++k) {
for (int l = 0; l <= 9; ++l) {
for (int p = 0; p <= 9; ++p) {
for (int q = 0; q <= 9; ++q) {
sum[i][j][k][l][p][q] += sum[i][j - 1][k][l][p][q];
}
}
}
}
}
}
for (int i = 1; i <= 9; ++i) {
for (int j = 0; j <= 9; ++j) {
for (int k = 0; k <= 9; ++k) {
for (int l = 0; l <= 9; ++l) {
for (int p = 0; p <= 9; ++p) {
for (int q = 0; q <= 9; ++q) {
sum[i][j][k][l][p][q] += sum[i - 1][j][k][l][p][q];
}
}
}
}
}
}
ll ans = 0;
for (int i = 1; i <= n; ++i) {
ans += sum[9 - b[i][0]][9 - b[i][1]][9 - b[i][2]][9 - b[i][3]][9 - b[i][4]][9 - b[i][5]];
bool flag = 1;
for (int j = 0; j < 6; ++j) {
if (b[i][j] > 9 - b[i][j]) {
flag = 0;
break;
}
}
if (flag) {
--ans;
}
}
printf("%lld\n", ans >> 1);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}