1209: Three Jugs

1209: Three Jugs

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 245   Solved: 37
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Description

    We have three jugs A, B, C without any calibration, and an infinite supply of water. There are three types of actions that you can use:
     (1) Fill a jug.
     (2) Empty a jug.
     (3) Pour from one jug to another.
    Pouring from one jug to another stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons, B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A.
    Now you need to calculate the minimum accurate gallons of water we can get by using the three jugs.

Input

    There is an integer T (1 <= T <= 200) in the first line, means there are T test cases in total.
    For each test case, there are three integers a, b, c (1 <= a, b, c <= 10^18) in a line, indicate the capacity (unit: gallon) of the three jugs.

Output

    For each test case, you should print one integer in a line, indicates the minimum accurate gallons of water we can get by using the three jugs.

Sample Input

2
3 6 9
6 10 15

Sample Output

3
1

HINT

Source

CSU Monthly 2013 Apr.

 

 

找3个数的最大公约数

#include<stdio.h>

unsigned long   long   gcd(unsigned long   long   a,unsigned long   long   b)

{

if(a==0)

return   b;

else

return   gcd(b%a,a);

}

int   main()

{

int   t;

unsigned long   long   a,b,c,ans;//记得改回long long

while(scanf("%d",&t)!=EOF)

{

while(t--)

{

scanf("%lld %lld %lld",&a,&b,&c);

ans=gcd(a,gcd(b,c));

printf("%lld\n",ans);

}

}

return   0;

}