经典算法问题之打印日期

这也是一道经典的算法题。

其实也是用两个数组。还有判断是否闰年。

两个个循环，外面一个是月份循环，内部一个是每个月的天数循环，然后计数器Count++就行，直到和天数相同就跳出循环，打印就行。

#include <stdio.h>
int judge ( int year ) {
    if (year % 400 == 0 || year % 100 != 0 && year % 4 == 0 ) {
        return 1 ;
    }
    return 0 ;
}
int main ( ) {
    int year , days ;
    int A[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    int B[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
    while( scanf("%d %d",&year,&days) != EOF ) {
        int i = 1 , tem = 0 ;
        int count = 0 ;
        if( judge(year) == 1 ) {
            for ( ; i <= 12 ; i++ ) {
                for( int j = 1 ; j <= B[i] ; j++ ) {
                    count++;
                    if( count == days ) {
                        tem = j ;
                        break ;
                    }
                }
                if( count == days ) {
                    break ;
                }
            }
        }else{
            for ( ; i <= 12 ; i++ ) {
                for( int j = 1 ; j <= A[i] ; j++ ) {
                    count++;
                    if( count == days ) {
                        tem = j ;
                        break ;
                    }
                }
                if( count == days ) {
                    break ;
                }
            }
        }
        if( i < 10 ) {
            if ( tem < 10 ) {
                printf( "%d-0%d-0%d\n",year,i,tem ) ;
            }else {
                printf( "%d-0%d-%d\n",year,i,tem ) ;
            }
        }else{
            if ( tem < 10 ) {
                printf( "%d-%d-0%d\n",year,i,tem ) ;
            }else {
                printf( "%d-%d-%d\n",year,i,tem ) ;
            }
        }
    }
    return 0 ;
}

结果如下：