【HDU】2669 - Romantic(扩展欧几里得)
2016-08-01 17:30
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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4998 Accepted Submission(s): 2078
Problem Description
The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei
Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.
Input
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)
Output
output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.
Sample Input
77 51
10 44
34 79
Sample Output
2 -3
sorry
7 -3
Author
yifenfei
Source
HDU女生专场公开赛——谁说女子不如男
全裸的一道扩展欧几里得题。注意用 __int64 。
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
__int64 exGCD(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
if (!b)
{
x = 1;
y = 0;
return a;
}
__int64 g = exGCD(b,a%b,y,x);
y -= a / b * x;
return g;
}
int main()
{
__int64 a,b,c,d,x,y;
while (~scanf ("%I64d %I64d",&a,&b))
{
c = exGCD(a,b,x,y);
if (c != 1)
puts("sorry");
else
{
x = (x % b + b) % b;
printf ("%I64d %I64d\n",x,(1-a*x) / b);
}
}
return 0;
}
Romantic
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4998 Accepted Submission(s): 2078
Problem Description
The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei
Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.
Input
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)
Output
output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.
Sample Input
77 51
10 44
34 79
Sample Output
2 -3
sorry
7 -3
Author
yifenfei
Source
HDU女生专场公开赛——谁说女子不如男
全裸的一道扩展欧几里得题。注意用 __int64 。
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
__int64 exGCD(__int64 a,__int64 b,__int64 &x,__int64 &y)
{
if (!b)
{
x = 1;
y = 0;
return a;
}
__int64 g = exGCD(b,a%b,y,x);
y -= a / b * x;
return g;
}
int main()
{
__int64 a,b,c,d,x,y;
while (~scanf ("%I64d %I64d",&a,&b))
{
c = exGCD(a,b,x,y);
if (c != 1)
puts("sorry");
else
{
x = (x % b + b) % b;
printf ("%I64d %I64d\n",x,(1-a*x) / b);
}
}
return 0;
}
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