GYM102082 A Digits Are Not Just Characters

https://codeforces.com/gym/102082

题意:按照它的规则来判断字符串的大小。

分析:按照题意模拟即可。

#include "bits/stdc++.h"

namespace fastIO {
#define BUF_SIZE 100000
    bool IOerror = 0;

    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if (p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if (pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }

    inline bool blank(char ch) { return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t'; }

    inline void read(long long &x) {
        char ch;
        while (blank(ch = nc()));
        if (IOerror) return;
        for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }

#undef BUF_SIZE
};
using namespace fastIO;
using namespace std;

const int mod = 1e9 + 7;

int cmp(string s, string t) {
    if ((s[0] >= '0' && s[0] <= '9') && ((t[0] >= '0' && t[0] <= '9'))) {
        int pos1 = s.length() - 1, pos2 = t.length() - 1;
        if (pos1 < pos2)return -1;
        else if (pos1 > pos2)return 1;
        else {
            pos1 = pos2 = 0;
            while (pos1 < s.length()) {
                if (s[pos1] < t[pos2])return -1;
                else if (s[pos1] > t[pos2])return 1;
                pos1++;
                pos2++;
            }
            return 0;
        }
    } else if ((s[0] >= '0' && s[0] <= '9') && ((t[0] < '0' || t[0] > '9'))) {
        return -1;
    } else if ((s[0] < '0' || s[0] > '9') && ((t[0] >= '0' && t[0] <= '9'))) {
        return 1;
    } else {
        return s.compare(t);
    }
}

int main() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    while (n--) {
        string t;
        cin >> t;
        int pos1 = 0, pos2 = 0;
        int ans;
        while (1) {
            string s1 = "", s2 = "";
            if (pos1 == s.length() && pos2 == t.length()) {
                ans = 0;
                break;
            } else if (pos1 == s.length() && pos2 < t.length()) {
                ans = -1;
                break;
            } else if (pos1 < s.length() && pos2 == t.length()) {
                ans = 1;
                break;
            }
            while (pos1 < s.length() && (s[pos1] >= '0' && s[pos1] <= '9'))s1 += s[pos1++];
            while (pos2 < t.length() && (t[pos2] >= '0' && t[pos2] <= '9'))s2 += t[pos2++];
            if (s1 == "")s1 += s[pos1++];
            if (s2 == "")s2 += t[pos2++];
            int x = cmp(s1, s2);
            if (x == 0)continue;
            else if (x == 1) {
                ans = 1;
                break;
            } else if (x == -1) {
                ans = -1;
                break;
            }
        }
        if (ans == 1)puts("-");
        else puts("+");
    }
}

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转载自blog.csdn.net/qq_42671946/article/details/89842853
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