解法：直接上模版。
扩展欧几里德的模版：
typedef long long LL;
LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
if(b==0)
{
x=1;
y=0;
return a;
}
LL d=ex_gcd(b,a%b,x,y);
LL t=x;
x=y;
y=t-a/b*y;
return d;
}
求解一元线性同余方程组模版：
LL solve(LL n)
{
LL a1,r1,a2,r2;
LL a,b,c,r,x,y;
bool ifhave=true;
scanf("%lld%lld",&a1,&r1);
for(LL i=1;i
题目： poj 2891 Strange Way to Express Integers
代码：
#include
#include
#include
using namespace std;
typedef long long LL;
LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
if(b==0)
{
x=1;
y=0;
return a;
}
LL d=ex_gcd(b,a%b,x,y);
LL t=x;
x=y;
y=t-a/b*y;
return d;
}
LL solve(LL n)
{
LL a1,r1,a2,r2;
LL a,b,c,r,x,y;
bool ifhave=true;
scanf("%lld%lld",&a1,&r1);
for(LL i=1;i>n)
{
cout<
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