cf/Codeforces ECR16-D - Two Arithmetic Progressions-中国剩余定理

Problem ECR16-D - Two Arithmetic Progressions-

题目大意

 给两个等差数列 You are
given two arithmetic progressions: a1k + b1 and a2l + b2.
Find the number of integers x such
that L ≤ x ≤ R andx = a1k' + b1 = a2l' + b2,
for some integers k', l' ≥ 0.
 

给一个L,R，问L到R里有多少个x满足  x = a1k' + b1 = a2l' + b2,

 　　数据范围：a1,a2,b1,b2,L，R <= 10^18。

解题分析

　　 其实题目的式子可以转化为两个模仿方程，也就是 x=b1(mod a1 )  ，x=b2(mod a2 )

由于a1，a2并不互质，不能直接用中国剩余定理，我们需要通过把式子都合并起来，再使用中国剩余定理

 
如果两个式子不能合并，那么就是找不到x，输出0，否则就找一个比L小的 正解cur，然后看看cur到L-1里有多少个解，cur到R里有多少个解。

由于题目的直线斜率是正的，因此 x的最小取值为max(b1,b2)，L取max(L,max(b1,b2))

 

参考程序



 #include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;
typedef long long ll;
const int N = 1005;

ll a
, m
;

ll gcd(ll a,ll b)
{
return b? gcd(b, a % b) : a;
}

void extend_Euclid(ll a, ll b, ll &x, ll &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
extend_Euclid(b, a % b, x, y);
ll tmp = x;
x = y;
y = tmp - (a / b) * y;
}

ll Inv(ll a, ll b)
{
ll d = gcd(a, b);
if(d != 1) return -1;
ll x, y;
extend_Euclid(a, b, x, y);
return (x % b + b) % b;
}

bool merge(ll a1, ll m1, ll a2, ll m2, ll &a3, ll &m3)
{
ll d = gcd(m1, m2);
ll c = a2 - a1;
if(c % d) return false;
c = (c % m2 + m2) % m2;
m1 /= d;
m2 /= d;
c /= d;
c *= Inv(m1, m2);
c %= m2;
c *= m1 * d;
c += a1;
m3 = m1 * m2 * d;
a3 = (c % m3 + m3) % m3;
return true;
}
ll cur,mm;
ll CRT(ll a[], ll m[], int n)
{
ll a1 = a[1];
ll m1 = m[1];
for(int i=2; i<=n; i++)
{
ll a2 = a[i];
ll m2 = m[i];
ll m3, a3;
if(!merge(a1, m1, a2, m2, a3, m3))
return -1;
a1 = a3;
m1 = m3;
}
cur=a1,mm=m1;
return (a1 % m1 + m1) % m1;
}

int main()
{
int n=2;
ll L,R;
for(int i=1; i<=n; i++)
scanf("%lld%lld",&m[i], &a[i]);
scanf("%lld%lld",&L,&R);
ll ans = CRT(a, m, n);
if (ans==-1)
{
printf("0\n");
return 0;
}
L=max(L,max(a[1],a[2])); //***
//cal ans between [L-1,R]
ans=0;
if (cur>=L) cur-=((cur-L)/mm+1)*mm; //find a cur that < L
if (L<=R)
ans= (R-cur)/mm-(L-1-cur)/mm;

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

return 0;
}