HihoCoder 1871 B. Heshen's Account Book

离谱的语文题

5点的时候我说，这不是傻逼题吗，我来打，最后看了题解之后八点多才做出来，太离谱了

1abc111abc1可以提炼出11111这个数字（这谁看的出来呀）

#include <bits/stdc++.h>
using namespace std;
const int N = 2 * 1e6 + 10;
char s[N];
int ans[N], h[N];
#define isdight(x) (x >= '0' && x <= '9')
int main() {
    int p = 0, last = 0, cnt = 1, last_p = 0;
    while (scanf("%[^\n]", &s[p]) != EOF) {
        getchar();
        p = strlen(s);
        s[p++] = ' ';
        for (int i = last; i < p; i++)
            h[i] = cnt;
        cnt++; char c;                                                                             
        if ((c = getchar()) == EOF)    break;
        if (isdigit(c) && isdigit(s[p - 2]))
            p--;
        last = p;
        s[p++] = c; last_p = p;
        h[p - 1] = cnt;
    }
    if (last_p == p)    cnt++, s[p++] = ' ';
    int len = strlen(s);
    int pos = 0;
    for (int i = 0; i <= len; i++) {
        if (s[i] == ' ' || s[i] == '\r' || s[i] == '\n') {
            if (!isdigit(s[pos]) || !isdigit(s[i - 1]));
            else if (s[pos] == '0') {
                if (i - pos == 1) {
                    ans[h[pos]]++;
                    printf("0 ");
                }
            }
            else {
                for (int j = pos; j < i; j++)
                    if (isdigit(s[j]))
                        printf("%c", s[j]);
                if (pos < i)    printf(" "), ans[h[pos]]++;
            }
            pos = i + 1;
        }
        else {
            pos = min(pos, i);
        }
    }
    puts("");
    for (int i = 1; i < cnt; i++)
        printf("%d\n", ans[i]);
    return 0;
}