2015
05-23

# Enumeration?

In mathematics and theoretical computer science, the broadest and most abstract definition of an enumeration of a set is an exact listing of all of its elements (perhaps with repetition).
If we are blocked by a tricky but small issue, a good way is stop bothering and calculating, and just enumerates all the possible result. Here is such a case. iSea is fond of coins recently, now he gets three coins, after throwing all of them again and again, he finds some of those coins are not side equivalent, because the times it faces up and faces down are not equal. Here we assume iSea has thrown the coin enough times that we can treat this as random happened and use the number to count probability.
iSea is a typical Libra, placing in equilibrium at the first position always. He want to distribute the probability to the three coins, and each time he chooses a coin from them with this probability, throws the coin, records facing up or down, and this time after summing up all the times whether the coins face up or down the ideal probable number should be perfectly equal. Again, is this possible? Remember the probability for each coin can’t be zero.

The first line contains a single integer T, indicating the number of test cases.
Each test case includes Six integers Up1, Down1, Up2, Down2, Up3, Down3, indicating the times facing up and down of the first, second and third coin, respectively.

Technical Specification
1. 1 <= T <= 10 000
2. 1 000 000 <= Up, Down <= 2 000 000

The first line contains a single integer T, indicating the number of test cases.
Each test case includes Six integers Up1, Down1, Up2, Down2, Up3, Down3, indicating the times facing up and down of the first, second and third coin, respectively.

Technical Specification
1. 1 <= T <= 10 000
2. 1 000 000 <= Up, Down <= 2 000 000

2
1000000 1000001 1000000 1000002 1000000 1000003
1000000 1000001 1000000 1000002 2000000 1000003

Case 1: No
Case 2: Yes

1：如果三枚硬币正面朝上的概率都是50%，则可以达到要求

2：若果有硬币的概率大于50%，一定要有概率小于50%的才能抛出后达到50%

#include<stdio.h>
int main()
{
int i,a[3],b[3],c[3],t,op=1;
scanf("%d",&t);
while(t--)
{
for(i=0;i<3;i++)
c[i]=0;
for(i=0;i<3;i++)
{
scanf("%d%d",&a[i],&b[i]);
if(a[i]==b[i])
c[0]++;
else if(a[i]>b[i])
c[1]++;
else c[2]++;
}
printf("Case %d: ",op++);
if(c[0]==3||(c[1]>0&&c[2]>0))
puts("Yes");
else puts("No");
}
return 0;
}