请问这题该怎么做才能考虑到所有错误情况？不能用数组和函数

Question

Answer 1

#include<stdlib.h>
#include<stdio.h>
int main(){
    // 2021 2 1
    int y,m,weekday,state,totalDays,i,j,row,col,t;
    state=scanf("%d%d",&y,&m);
    while(state!=2||y<2000||y>2030||m<1||m>12){
        fflush(stdin);
        state=scanf("%d%d",&y,&m);
    }

    if(m==1||m==3||m==5||m==7||m==8||m==10||m==12)  totalDays = 31;
    else if(m==2)   totalDays =((y%4==0&&y%100!=0)||y%400==0)?29:28;
    else totalDays = 30;

    state=scanf("%d",&weekday);
    while(state!=1||weekday<0||weekday>6){
        fflush(stdin);
        state=scanf("%d",&weekday);
    }
    printf("日\t一\t二\t三\t四\t五\t六\n");
    row = (weekday + totalDays)/7+1;
    col = 7;t=0;
    for(i=0;i<row;i++){
        for(j=0;j<col;j++){
            if(i==0&&j<weekday){
                printf(" \t");
            }else if(t<totalDays){
                t += 1;
                printf("%d\t",t);
            }
        }
        printf("\n");
    }
    system("pause");
}

Answer 2

我学的是java，不过思路差不多，写个死循环，然后里边写判断，错误的话直接continue就行了，还得写个异常，异常来判断不是数字的类型，错误的话也continue重新循环