Codeforces 710D Two Arithmetic Progressions（同余方程组）-优快云博客

本文介绍了解同余方程组的方法，包括使用扩展欧几里得算法(exgcd)来寻找满足特定条件的解，并提供了两种不同的实现代码。通过这些方法可以有效地找到满足多个同余条件的解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

$x=a_1k+b_1=a_2l+b_2$
$x=b_1(mod a_1),x=b_2(mod a_2)$
$同余方程组，两个模数b_1,b_2不需要处理，直接扔进模版，L=max(max(b_1,b_2),L)$
$因为k,l\ge0，所以L要取最大$
$求出来x之后，lcm(a_1,a_2)的加，求区间里有多少个点就行了$
$这题无解模版里不能返回-1，因为可能解就是-1$

$方法2是直接用exgcd求解$
$a_1x+a_2y=b_2-b_1$
$求出x，然后化成x的最小正数解形式$
$然后求出a_1x+b1的最小解，然后L取b_1,b_2,L之间最大的$
$然后每步是lcm(a_1,a_2)$
$和上面一样求就行了$

代码

#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           10000005
#define   MAXN          1000005
#define   maxnode       205
#define   sigma_size    26
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-4;
const LL    mod    = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
 }
/*****************************************************/ 

vector<int> a,b;

LL mul(LL a,LL b,LL mod){
    LL n = 0;
    while(b){
        if(b&1)
          n = (n+a)%mod;
        a = (a*2)%mod;
        b /= 2;
    }
    return n;
}

void exgcd(LL a,LL b,LL &d,LL &x,LL &y){
    if(!b){x=1;y=0;d=a;return;}
    else {exgcd(b,a%b,d,y,x);y-=a/b*x;}
}

LL labs(LL a){
    if(a<0) return -a;
    return a;
}
LL solve(){ //x=b[i](mod a[i])
    LL ta=a[0],tb=b[0];
    int flag=1;
    for(int i=1;i<a.size();i++){
        LL xa=ta,xb=a[i],c=b[i]-tb,d,x,y;
        exgcd(xa,xb,d,x,y);
        if(c%d){
            flag=0;
            break;
        }
        LL tmp=xb/d;
        LL k=1;
        if(x<0) k*=-1;
        if(c/d<0) k*=-1;
        x=(mul(labs(x),labs(c/d),tmp)*k+tmp)%tmp;//加入了二分快速乘，需要把负数先变为正数
        tb=ta*x+tb; 
        ta=ta/d*a[i];
    }
    if(!flag) return -INF;
    return tb;
}

LL gcd(LL x,LL y){
    if(!y) return x;
    return gcd(y,x%y);
}

LL lcm(LL x,LL y){
    return x*y/gcd(x,y);
}
int main(){
    LL a1,b1,a2,b2,L,R;
    while(cin>>a1>>b1>>a2>>b2>>L>>R){
        a.clear();b.clear();
        a.push_back(a1);
        a.push_back(a2);
        b.push_back(b1);
        b.push_back(b2);
        LL cnt=solve();
        L=max(max(b1,b2),L);
        //cout<<cnt<<endl;
        if(L>R){
            cout<<0<<endl;
            continue;
        }
        if(cnt==-INF){
            cout<<0<<endl;
            continue;
        }
        LL tmp=lcm(a1,a2);
        LL ans=0;
        if(cnt<=R) ans+=(R-cnt)/tmp+1;
        if(cnt<L) ans-=(L-1-cnt)/tmp+1;
        cout<<ans<<endl;
    }
    return 0;
}

代码2：

#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           100005
#define   MAXN          1000005
#define   maxnode       205
#define   sigma_size    26
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-4;
const LL    mod    = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
 }
/*****************************************************/ 

void exgcd(LL a,LL b,LL &d,LL &x,LL &y){  
    if(!b){x=1;y=0;d=a;return;}  
    else {exgcd(b,a%b,d,y,x);y-=a/b*x;}  
}  

LL labs(LL a){
    if(a<0) return -a;
     return a;
}
int main(){
    LL a1,b1,a2,b2,L,R;
    while(cin>>a1>>b1>>a2>>b2>>L>>R){
        LL x,y,d;
        exgcd(a1,a2,d,x,y);
        if((b2-b1)%d!=0){
            cout<<0<<endl;
            continue;
        }
        x*=(b2-b1)/d;
        x=(x%labs(a2/d)+labs(a2/d))%labs(a2/d);
        LL cnt=x*a1+b1;
        LL tmp=labs(a1*a2/d);
        //cout<<cnt<<" "<<tmp<<endl;
        LL ans=0;
        L=max(L,max(b1,b2));
        if(L>R){
            cout<<0<<endl;
            continue;
        }
        if(cnt<=R) ans+=(R-cnt)/tmp+1;
        if(cnt<L) ans-=(L-1-cnt)/tmp+1;
        cout<<ans<<endl;
    }
    return 0;
}