海的那边是自由吗……………… 不 海的那边是

ikun 黄昏产生虚伪的拥护 落日见证真正的信徒 ikun永不落幕！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！

C++

折磨了我好久好久，已经是历史遗留产物了......

代码已经过处理，使冗长的代码变得清晰明了

#include <bits/stdc++.h>
using namespace std;

bool LeapYear(int y) { //闰年判断
    if((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) {
        return true;
    }
    else {
        return false;
    }
}

int Days(int y, int m) { //计算某年某月有几天
    if(m == 4 || m == 6 || m == 9 || m == 11) {
        return  30;
    }
    if(m == 2) {
        if(LeapYear(y)) return 29;
        else return  28;
    }
    else return  31;
}

void Print(int w, int d) { //输出
    cout << " Sun Mon Tue Wed Thu Fri Sat" << endl;
    if(w != 6) {
        for(int i = 0; i <= w; i++) cout << "    ";
    }
    int k;
    if(w == 6) k = -1;
    else k = w;
    for(int i = 1; i <= d; i++) {
        k += 1;
        if(k == 7) { //每一个星期一行
            cout << endl;
            k = 0;
        };
        printf("%4d", i);
    }
}

int main(){
    int y, m, d;
    cin >> y >> m;
    int w;

    d = Days(y, m);

    if(m == 1 || m == 2) { // 基姆拉尔森计算公式的坑点
        y -= 1;
        m += 12;
    }

    w = (1+2\*m+3\*(m+1)/5+y+y/4-y/100+y/400) % 7; //计算1日是星期几

    Print(w, d);
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int y,m,s,w,sm[13]={0,31,28,31,30,31,30,31,31,30,31,30,31},d[13][35],x;
int main(){
    cin>>y>>m;
    cout<<" Sun Mon Tue Wed Thu Fri Sat\n";
    s=y-1+((y-1)/4)-((y-1)/100)+((y-1)/400)+1;
    w=s%7;
    if((y%4==0&&y%100!=0)||y%400==0){
        sm[2]=29;
    }
    for(int i=1;i<=12;i++){
        for(int j=1;j<=sm[i];j++){
            d[i][j]=w;
            w++;
            if(w>6){
                w=0;
            }
            //cout<<s<<endl;
        }
    }
    for(int i=1;i<=d[m][1];i++){
        cout<<"    ";
    }
    for(int i=1;i<=sm[m];i++){
        if(i/10==0){
            cout<<' ';
        }
        cout<<"  "<<i;
        if(d[m][i]==6){
            cout<<endl;
        }
    }
    return 0;
}

C++ :

#include <iostream>
#include <cstdio>
using namespace std;
int days[12]={31,28,31,30,31,30,31,31,30,31,30,31};
int isleap(int year);
int main()
{
    int year,month;
    scanf("%d%d",&year,&month);
    int sum=0,k=1;//365%7
    for(int i=1; i<=year-1; i++){
        if(isleap(i)){
            sum+=2;
        }else{
            sum+=1;
        }
    }
    sum%=7;
    if(isleap(year)&& month>2){sum++;}
    for(int i=1; i<month; i++){
        sum+=days[i-1];
    }
    sum%=7;
    int day=days[month-1];
    if(isleap(year)&& month==2){day++;}
    printf(" Sun Mon Tue Wed Thu Fri Sat\n");
    if(sum<6){
        for(int i=0; i<=sum; i++){
            printf("    ");
        }
    }

    for(int i=1; i<=day; i++){
        printf("%4d",i);
        if(++sum%7==6){printf("\n");}
    }
    return 0;
}

int isleap(int year)
{
    if(year%400==0 || (year%4==0 && year%100!=0)){
        return 1;
    }else{
        return 0;
    }
}