\[1 \ 2 \ 2 \ 5 \\ \ \ 1 \ 2 \ 2 \\ \ \ \ \ 1 \ 2 \\ \ \ \ \ \ \ 1 \\ \]

观察发现每一位只需要维护后缀和

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 5e5 + 10;
int n;
LL sum[N];
char s[N];
int main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    vector<int> c;
    LL tot = 0;
    for (int i = 1; i <= n; i ++ ) sum[i] = sum[i - 1] + s[i] - '0';
    reverse(sum + 1, sum + 1 + n);
    for (int i = 1; i <= n; i ++ ) {
        tot += sum[i];
        c.push_back(tot % 10);
        tot /= 10;
    }
    while (tot) {
        c.push_back(tot % 10);
        tot /= 10;
    }
    reverse(c.begin(), c.end());
    for (auto x : c)
        printf("%d", x);
    printf("\n");
    return 0;
}