题解 | 日期累加

#include<iostream>
#include <iomanip>
using namespace std;

bool isLeap(int y) {
    if ((y % 400 == 0) || (y % 100 != 0 && y % 4 == 0)) {
        return true;
    }
    else {
        return false;
    }
}

int dayofyear(int y, int m, int d) {
    static const int days[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    int sum = d;
    for (int i = 0;i < m - 1;i++) {
        sum = sum + days[i];
    }
    if (m > 2 && isLeap(y)) {
        sum = sum + 1;
    }
    return sum;
}

void nthtodate(int y, int n, int& m, int& d) {
    // 每月天数表：普通年
    int days[12] =
    { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

    if (isLeap(y)) {
        days[1] = 29;
    }
    m = 1;
    while (n > days[m - 1]) {
        n = n - days[m - 1];
        m++;
    }
    d = n;
}

int main() {
    int x ,y, m, d, sum;
    cin >> x;
    while (cin >> y >> m >> d >> sum) {
        int ans = dayofyear(y, m, d);
        sum = sum + ans;

        if (isLeap(y) && sum > 366) {
            y = y + 1;
            sum = sum - 366;
        }
        if (!isLeap(y) && sum > 365) {
            y = y + 1;
            sum = sum - 365;
        }

        nthtodate(y, sum, m, d);

        cout << y << '-' << setw(2) << setfill('0') << m << '-' << setw(2) << setfill('0') << d << endl;
    }
    return 0;
}