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关于日期处理的常用方法，掌握闰年的计算方式和各月的天数即可满足大部分需求。

例题剖析

回文日期

2020年蓝桥杯A组

样例输入: 20200202

样例输出: 2021120221211212

代码解析: #include

int days[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; bool isLeap(int year) { return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0); } int getDay(int year, int month) { if (month == 2) return days[2] + isLeap(year); return days[month]; } int main() { int n; cin >> n; string ans1, ans2; bool flag1 = false, flag2 = false;

for (int i = n / 10000 + 1; i <= 9999; i++) {
    string a = to_string(i);
    string b = a;
    reverse(b.begin(), b.end());
    int month = stoi(b.substr(0, 2));
    int day = stoi(b.substr(2, 2));
    if (month < 1 || month > 12) continue;
    if (day < 1 || day > getDay(i, month)) continue;
    string s1 = a.substr(0, 2);
    string s2 = a.substr(2, 2);
    if (!flag1) {
        ans1 = a + b;
        flag1 = true;
    }
    if (!flag2 && s1 == s2 && s1[0] != s1[1]) {
        ans2 = a + b;
        flag2 = true;
    }
    if (flag1 && flag2) break;
}
cout << ans1 << endl;
cout << ans2 << endl;

}

跑步锻炼

2020年蓝桥杯B题

代码解析: #include

int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; bool isLeap(int year) { return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0); } int getDay(int year, int month) { if (month == 2) return days[2] + isLeap(year); return days[month]; }

int main() { int ans = 0, weekday = 4; for (int i = 2000; i < 2020; i++) { for (int j = 1; j <= 12; j++) { for (int k = 1; k <= getDay(i, j); k++) { weekday = (weekday + 1) % 7; if (k == 1 || weekday == 0) ans += 2; else ans++; } } } for (int j = 1; j < 10; j++) { for (int k = 1; k <= getDay(2020, j); k++) { weekday = (weekday + 1) % 7; if (k == 1 || weekday == 0) ans += 2; else ans++; } } ans += 2; // 2020.10.1 cout << ans << endl; }

答案: 8879

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