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  • 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("%[^
    ]", &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] == '
    ' || s[i] == '
    ') {
                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
    ", ans[i]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cminus/p/14093504.html
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