你看看我的程序！
char month2[20][10] ={ "imix ", "ik ", "akbal ", "kan ", "chicchan ", "cimi ", "manik ", "lamat ", "muluk ", "ok ", "chuen ","eb ", "ben ", "ix ", "mem ", "cib ", "caban ", "eznab ", "canac ", "ahau "};
这儿按初始数据来的，
后来的天数需要+1

#include <string.h>
#include <stdio.h> 
#define NUM_DAY1 365
#define NUM_DAY2 (13 * 20)
int main()
{
	char month1[19][9]={"pop", "no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac", "ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};
	char month2[20][10] ={ "imix ", "ik ", "akbal ", "kan ", "chicchan ", "cimi ", "manik ", "lamat ", "muluk ", "ok ", "chuen ","eb ", "ben ", "ix ", "mem ", "cib ", "caban ", "eznab ", "canac ", "ahau "};
	int year1,day1,year2,day2,day,n,i,cnt,r;
	char t,input[30],month[10],*nameOfDay=(char *)malloc(10*sizeof(char));
	scanf("%d",&n);
	printf("%d\n",n);
	while(n>0){
		scanf("%d",&day1);
		scanf("%c",&t);
		scanf("%s",month);
		scanf("%d",&year1);
		
		//change
		for (i=0;strcmp(month,month1[i]);i++);
		cnt = year1 * NUM_DAY1 + i * 20 + day1;		
		year2 = cnt / NUM_DAY2;
		r = cnt % NUM_DAY2;
		day2 = r % 13 + 1;
		nameOfDay = month2[r % 20];
		printf("%d %s%d\n",day2,nameOfDay,year2);
		n--;
 	}
	return 0;
}

• Re:测试数据也过了 但是总是不能通过 (1353) kevin655 2006-04-03 19:59:09