第三章：数据结构决定程序

1.第1题解答

程序是线性分段函数，可用二分法定位对应的分段函数。

2.第4题解答

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int month[13] = {
0,    31,    28,    31,    30,    31,    30,    31,    31,    30,    31,    30,    31
//    1    2    3    4    5    6    7    8    9    10    11    12
};

class D{
public:
    int year,mon,day;// 1900 <= year, 1 <= mon <= 12,
                    // 1 <= day <= 31
    D(){}
    D(int y, int m, int d):
        year(y),mon(m),day(d){}

    int yearday(void){//返回这一天是这一年的第几天
        int sum = day;
        for( int i = 1; i < mon; i++ )
            sum += month[i];
        if( isrun() && mon > 2 )
            sum ++;
        return sum;
    }
    bool isrun( void ){//是否是闰年
        return (year%4==0&&year%100!=0)||(year%400==0);
    }
};

int dist( D d1, D d2 ){//两个日期相差的天数
    int sum = -(d1.yearday());
    for( ; d1.year < d2.year ; d1.year++ )
        sum += d1.isrun()?366:365;
    return sum + d2.yearday();
}

int xingqiji( D d ){//某一天是星期几
    D temp(1900,1,1);
    return dist( temp, d )%7+1;
}

int print(int year, int mon ){//输出某月日历
    D d(year, mon, 1 );
    int week = xingqiji(d);
    int sum = month[ mon ];
    for( int i = 1; i < week; i++ )
        cout << "    ";
    for( int i = 1; i <= sum; i++){
        cout << i << "    ";
        if( week == 7 ){
            week = 1;
            cout << endl;
        }
        else week++;
    }
    cout << endl;
}

int main(void){
    D a(2015,3,5);
    cout << xingqiji(a) << endl;
    print(2015,3);
    return 0;
}

3.第8题解答

(待补充...)