1008. 二哥买期货

http://acm.sjtu.edu.cn/OnlineJudge/problem/1008

对起始年份和结束年份，可以对每一天单独判断；

对中间的每个整年，周末总的天数分为两部分：

1. 每个整年恰有完整的52个周，所以至少有 2*52 天是周末；

2. 闰年时，366%7 = 2，需要判断12-31和12-30是否为周末即可，平年需要判断12-31是否为周末；

对中间的每个整年，都有11天假日，注意周末与假日重合的情况。

这道题麻烦之处在于星期的计算和周末与假日重合时的处理。

星期的计算公式：

int getweek(int y, int m, int d)
{
    if (m==1 || m==2) m += 12, y -= 1;
    return (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400+1) % 7;
}

单独对每个假日进行判断，就避免了周末与假日重合时重复计算：

int holidays[11][2] = {{1,1}, {5,1}, {5,2}, {5,3}, {10,1}, {10,2}, {10,3}, {10,4}, {10,5}, {10,6}, {10,7}};

for (int j = 0; j < 11; ++j) {
    if (isWeekend(curYear, holidays[j][0], holidays[j][1]))
        ++ans;
}

(枚举的做法会超时，不过需要对拍时很有用。)

# include <stdio.h>

int holidays[11][2] = {{1,1}, {5,1}, {5,2}, {5,3}, {10,1}, {10,2}, {10,3}, {10,4}, {10,5}, {10,6}, {10,7}};

int getweek(int y, int m, int d)
{
    if (m==1 || m==2) m += 12, y -= 1;
    return (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400+1) % 7;
}
bool isWeekend(int y, int m, int d)
{
    int w = getweek(y, m, d);
    return (w==0 || w==6);
}

const int days[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool isLeapYear(int year) { return year%400==0 || (year%4==0 && year%100!=0); }

void getdate(int &y, int &m, int &d)
{
    char s[20];
    scanf("%s", s);
    y = (s[0]-'0')*1000+(s[1]-'0')*100+(s[2]-'0')*10+(s[3]-'0');
    m = (s[5]-'0')*10+(s[6]-'0');
    d = (s[8]-'0')*10+(s[9]-'0');
}

bool isHoliday(int m, int d)
{
    for (int i = 0; i < 11; ++i) {
        if (holidays[i][0] == m && holidays[i][1] == d) return true;
    }
    return false;
}

bool isEndOfMonth(int y, int m, int d)
{
    if (m != 2 && d == days[m]) return true;
    else if (m==2) {
        if (isLeapYear(y)) return d == 29;
        else return d == 28;
    }
    return false;
}

int cal(int year, int m, int d, int mm, int dd)
{
    int cnt = 0;
    for (int i = m, j = d; i < mm || (i==mm && j<=dd); ) {
        if (!isWeekend(year, i, j) && !isHoliday(i, j)) {
            ++cnt;
        }
        if (isEndOfMonth(year, i, j)) {
            ++i, j = 1;
        } else {
            ++j;
        }
    }
    return cnt;
}

void solve(void)
{
    int n;
    int y1, m1, d1;
    int y2, m2, d2;
    scanf("%d", &n);
    while (n--) {
        getdate(y1, m1, d1);
        getdate(y2, m2, d2);

        int ans = 0;

        if (y1 == y2) {
            ans = cal(y1, m1, d1, m2, d2);
        } else {
            ans = cal(y1, m1, d1, 12, 31) + cal(y2, 1, 1, m2, d2);
            for (int i = y1+1; i < y2; ++i) {
                if (isLeapYear(i)) {
                    ans += 366-52*2;
                    if (isWeekend(i, 12, 30)) --ans;
                    if (isWeekend(i, 12, 31)) --ans;
                } else {
                    ans += 365-52*2;
                    if (isWeekend(i, 12, 31)) --ans;
                }
                ans -= 11;
                for (int j = 0; j < 11; ++j) {
                    if (isWeekend(i, holidays[j][0], holidays[j][1]))
                        ++ans;
                }
            }
        }

        printf("%d
", ans);
    }
}

int main()
{
    solve();

    return 0;
}