Codeforces 710D Two Arithmetic Progressions（同余方程组）

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x=a1k+b1=a2l+b2
x=b1(moda1),x=b2(moda2)
同余方程组，两个模数b1,b2不需要处理，直接扔进模版，L=max(max(b1,b2),L)
因为k,l≥0，所以L要取最大
求出来x之后，lcm(a1,a2)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
a1x+a2y=b2−b1
求出x，然后化成x的最小正数解形式
然后求出a1x+b1的最小解，然后L取b1,b2,L之间最大的
然后每步是lcm(a1,a2)
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;}

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