题解 | 不要三句号的歪

#include <iostream>
#include<vector>

using namespace std;
#define int long long
vector<int> sub(vector<int>& A, vector<int>& B) {
    vector<int> C;
    for (int i = 0, t = 0; i < A.size(); i ++ ) {
        t = A[i] - t;
        if (i < B.size()) t -= B[i];
        C.push_back((t + 10) % 10);
        if (t < 0) t = 1;
        else t = 0;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

signed main() {
    string s;
    cin >> s;
    int i = 0;
    vector<int> a, b;
    int res1 = 0, res2 = 0, res3 = 0;
    while (s[i] != ',') {
        res1 = res1 * 10 + s[i] - '0';
        i++;

    }
    i++;
    while (s[i] != ',') {
        res2 = res2 * 10 + s[i] - '0';
        i++;
        a.push_back(s[i]);
    }
    i++;
    while (s[i] == '.') {
        i++;

    }
    i++;
    while (s[i] >= '0' && s[i] <= '9' && i < s.length()) {
        res3 = res3 * 10 + s[i] - '0';
        // b.push_back(s[i]);
        i++;

    }

    cout<<res3-res2-1;
    return 0;
}
// 64 位输出请用 printf("%lld")

本来是想用高精度，但是只有20位，用longlong就可以了，判断第三个数字和第二个数字，输出第三个数字-第二个数字-1即可。

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