poj 2115 简单数论题,同余方程/*poj 2115*/#includeusing namespace std;long long d;void ex_euclid(long long a,long long b,long long &x,long long &y){\x05if(b==0)\x05{\x05\x05x=1;\x05\x05y=0;\x05\x05d=a;\x05\x05return;\x05}\x05ex_euclid(b,a%b,x,
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![poj 2115 简单数论题,同余方程/*poj 2115*/#includeusing namespace std;long long d;void ex_euclid(long long a,long long b,long long &x,long long &y){\x05if(b==0)\x05{\x05\x05x=1;\x05\x05y=0;\x05\x05d=a;\x05\x05return;\x05}\x05ex_euclid(b,a%b,x,](/uploads/image/z/12571423-7-3.jpg?t=poj+2115+%E7%AE%80%E5%8D%95%E6%95%B0%E8%AE%BA%E9%A2%98%2C%E5%90%8C%E4%BD%99%E6%96%B9%E7%A8%8B%2F%2Apoj+2115%2A%2F%23includeusing+namespace+std%3Blong+long+d%3Bvoid+ex_euclid%28long+long+a%2Clong+long+b%2Clong+long+%26x%2Clong+long+%26y%29%7B%5Cx05if%28b%3D%3D0%29%5Cx05%7B%5Cx05%5Cx05x%3D1%3B%5Cx05%5Cx05y%3D0%3B%5Cx05%5Cx05d%3Da%3B%5Cx05%5Cx05return%3B%5Cx05%7D%5Cx05ex_euclid%28b%2Ca%25b%2Cx%2C)
poj 2115 简单数论题,同余方程/*poj 2115*/#includeusing namespace std;long long d;void ex_euclid(long long a,long long b,long long &x,long long &y){\x05if(b==0)\x05{\x05\x05x=1;\x05\x05y=0;\x05\x05d=a;\x05\x05return;\x05}\x05ex_euclid(b,a%b,x,
poj 2115 简单数论题,同余方程
/*poj 2115*/
#include
using namespace std;
long long d;
void ex_euclid(long long a,long long b,long long &x,long long &y)
{
\x05if(b==0)
\x05{
\x05\x05x=1;
\x05\x05y=0;
\x05\x05d=a;
\x05\x05return;
\x05}
\x05ex_euclid(b,a%b,x,y);
\x05long long t=x;
\x05x=y;
\x05y=t-(a/b)*y;
}
/* c * x = b - a mod (2 ^ k) */
int main()
{
\x05long long a,b,x,c,y;
\x05int k;
\x05while(cin>>a>>b>>c>>k)
\x05{
\x05\x05if (a == 0 && b == 0 && c == 0 && k == 0)
break;
\x05\x05long long m=(1
poj 2115 简单数论题,同余方程/*poj 2115*/#includeusing namespace std;long long d;void ex_euclid(long long a,long long b,long long &x,long long &y){\x05if(b==0)\x05{\x05\x05x=1;\x05\x05y=0;\x05\x05d=a;\x05\x05return;\x05}\x05ex_euclid(b,a%b,x,
有个细节你没注意到,看到范围是1