B Have a Nice Day

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Problem B: B Have a Nice Day

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 136   Solved: 40
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Description

Problem 

        Rumour has it that the P versus N P  question has been solved:  the two classes are not equal.  This 

implies  that  many  well-known  problems,  such  as  the  Traveling  Salesman  Problem,  will  remain 

difficult  forever.   It  can  be  considered  a  waste  of  time  to  search  for  polynomial  time  solutions: 

essentially only brute-force approaches can solve them.  Nothing you can do about that. 

    In  view  of  the  international  crisis,  the  new  Dutch  government  has  therefore  announced  that 

on  certain  days  it  is  not  allowed  to  work  on  these  hard  problems  anymore.  Instead,  one  must 

concentrate  on  easier  issues.   These  days  are  called  nice.   Of  course,  the  algorithm  to  decide 

whether a given date is nice or not should itself be easy.  So far politicians could not find such an 

algorithm.  Can you? 

    A  date  day month year is  written  down  using  the  digits  0,. . . ,9. A  date  is  called  nice if  the 

digits occurring in it occur an equal number of times,  and if it can be  split.       A date can be split 

if its four number set can be divided into two disjoint subsets with equal sum; the four numbers 

are the day, the month, the left part of the year (the number represented by its first and second 

digit; for 1957 this is 19) and the right part of the year (the number represented by its third and 

fourth  digit;  for  2000  this  is  0). For  example,  16  5  4928  is  nice,  because  all  digits  occur  exactly 

once and  16 + 5 + 28 = 49. 

Input

        The first line of the input contains a single number:  the number of test cases to follow.  Each test 

case has the following format: 

    •  One line with three integers D, M and Y  separated by single spaces, satisfying 1 ≤ D ≤ 31, 

       1 ≤ M ≤ 12 and  1000 ≤ Y  ≤ 9999:        the day,  month and year of a valid date,  respectively. 

Output

       For  every  test  case  in  the  input,  the  output  should  contain  the  string "yes" or  "no":  the  fact 

whether the date is nice or not. 

Sample Input

4
16 5 4928
14 12 2747
11 11 1111
3 3 2014

Sample Output

yes
no
yes
no

HINT

此题昨晚甚是纠结。。。。。。今天看了大神的代码，恍然大悟！！！！！！

只怪自己太弱。。。。

题意：给你三个数：day，month，year（左右各取两位，分成左右两个数）要求这四个数出现的次数相等，而且任意两个数相加等于另外两个数相加或者任意三个数相加等于另外一个数。

#include<stdio.h>
#include<string.h>
int digs[10];
void countdigs(int x)
{
	while(x)
	{
		x%=10;
		digs[x]++;
		x/=10;
	}
}
bool solve(int a,int b,int c,int d)
{
	return a == b + c + d || a + b == c + d || a + c == b + d || a + d == b + c || a + b + c == d || a + b + d == c || a + c + d == b;
}
int main()
{
	int t,d,m,y;
	scanf("%d",&t);
	while(t--)
	{
		memset(digs,0,sizeof(digs));
		scanf("%d%d%d",&d,&m,&y);
		countdigs(d);
		countdigs(m);
		countdigs(y);
		for(int i=0;i<10;i++)
		for(int j=0;j<10;j++)
		{
			if(digs[i]>0&&digs[j]>0&&digs[i]!=digs[j])
			{
				printf("no\n");
				return 0;
			}
		}
		if(solve(d,m,y/100,y%100))
		printf("yes\n");
		else printf("no\n");	
	}
	return 0;
}