首先可以列出线性方程组
方程组转化为在模p意义下的同余方程
因为不保证pp 互素，考虑扩展中国剩余定理合并
方程组是带系数的，我们要做的是在%p意义下把系数除过去,（系数为atk[i])
(atk[i],p[i]) 不等于1时无逆元，此时仍可能有解
很显然无解的情况就是
瞎jb猜的，无解的话就是%p[i]意义下atk[i] != 0 ,a[i] = 0
考虑原方程式ai = atk{i] * x + p[i] * y
方程两边同除gcd（pi,atki）解不变 此时atki,pi此时保证了atki与pi互素
若ai不能被整除也是无解的
扩展crt
着实被winXP下输出lld坑了一把....顺带被自己抄的splay坑了一把
/* 苟活者在淡红的血色中，会依稀看到微茫的希望 */ #include<bits/stdc++.h> using namespace std; inline long long read() { long long x; scanf("%lld",&x); return x; } int n,m; #define LL long long const int maxn = 500007; LL a[maxn],p[maxn]; // x atk[i] = a[i] ( % p[i]) LL atk[maxn],tatk[maxn]; struct Splay { #define fa(x) T[x].fa #define ls(x) T[x].ch[0] #define rs(x) T[x].ch[1] #define root T[0].ch[1] struct node { LL val,rev,siz,fa,ch[2]; }T[maxn]; LL tot; void clear() { tot = 0; root = 0; for(LL i = 1; i <= maxn; i++) T[i].val = T[i].rev = T[i].siz = T[i].fa = T[i].ch[0] = T[i].ch[1] = 0; } LL ident(LL x){return T[fa(x)].ch[0]==x?0:1;} void connect(LL x,LL fa,LL how){T[fa].ch[how]=x;T[x].fa=fa;} void update(LL x){T[x].siz=T[ls(x)].siz+T[rs(x)].siz+T[x].rev;} void rotate(LL x) { LL Y=T[x].fa,R=T[Y].fa; LL Yson=ident(x),Rson=ident(Y); LL B=T[x].ch[Yson^1]; connect(B,Y,Yson); connect(Y,x,Yson^1); connect(x,R,Rson); update(Y);update(x); } void splay(LL x,LL to) { to=T[to].fa; while(T[x].fa!=to) { if(T[fa(x)].fa==to) rotate(x); else if(ident(x)==ident(fa(x))) rotate(fa(x)),rotate(x); else rotate(x),rotate(x); } } LL newnode(LL fa,LL val) { T[++tot].fa=fa; T[tot].val=val; T[tot].rev=T[tot].siz=1; return tot; } LL find(LL val) { LL now=root; while(1) { if(T[now].val==val) {splay(now,root);return now;} LL nxt=T[now].val<val; now=T[now].ch[nxt]; } } void insert(LL val) { if(root==0) {root=newnode(0,val);return ;} LL now=root; while(1) { T[now].siz++; if(T[now].val==val) {T[now].rev++;splay(now,root);return ;} LL nxt=val<T[now].val?0:1; if(!T[now].ch[nxt]) {T[now].ch[nxt]=newnode(now,val);splay(now,root);return ;} now=T[now].ch[nxt]; } } void erase(LL val) { LL now=find(val); if(T[now].rev>1) {T[now].rev--;T[now].siz--;return ;} else if(!ls(now)&&!rs(now)) {root=0;return ;} else if(!ls(now)) {root=rs(now);T[rs(now)].fa=0;return ;} LL left=ls(now); while(rs(left)) left=rs(left); splay(left,ls(now)); connect(rs(now),left,1); connect(left,0,1); //update(rs(now)); update(left);// } LL pre(LL val) { LL now=root,ans=-1e13; while(now) { if(T[now].val<=val) ans=max(ans,T[now].val); LL nxt=val<=T[now].val?0:1; now=T[now].ch[nxt]; } return ans == -1e13 ? -1 : ans; } LL nxt(LL val) { LL now=root,ans=1e13; while(now) { if(T[now].val>val) ans=min(ans,T[now].val); LL nxt=val<T[now].val?0:1; now=T[now].ch[nxt]; } return ans; } }Sp; LL gcd(LL a,LL b) {return b == 0 ? a : gcd(b,a % b);} LL exgcd(LL a,LL b,LL &x,LL &y) { if(b == 0) {x = 1,y = 0;return a; } LL ret = exgcd(b,a % b,x,y); LL tmp = x;x = y;y = tmp - (a / b) * y; return ret; } LL inv(LL a,LL b) { LL x,y; exgcd(a,b,x,y); while(x < 0) x += b; return x; } void work() { LL ans = 0; for(int i = 1;i <= n;++ i) { ans = std::max(ans,a[i] % atk[i] == 0 ? a[i] / atk[i] : a[i] / atk[i] + 1); } printf("%lld\n",ans); } LL M[maxn],C[maxn]; inline LL add(LL x,LL y,LL mod) { return x + y >= mod ? x + y - mod : x + y; } LL mul(LL x,LL k,LL mod) { LL ret = 0 ; x %= mod; for(;k;k >>= 1,x = add(x,x,mod)) if(k & 1) ret = add(ret,x,mod); return ret; } bool flag = false; void init() { Sp.clear(); n = read(),m = read(); flag = false; //puts("haha"); for(int i = 1;i <= n;++ i) a[i] = read(); for(int i = 1;i <= n;++ i){ p[i] = read();if(p[i] != 1) flag = true; } for(int j = 1;j <= n;++ j) tatk[j] = read(); //printf("%d %d %d\n",n,m,flag); for(int k,i = 1;i <= m;++ i) k = read(),Sp.insert(k); for(int i = 1;i <= n;++ i) { //puts("asdasd"); LL p = Sp.pre(a[i]); if(p == -1) p = Sp.nxt(a[i]); atk[i] = p; Sp.erase(p); Sp.insert(tatk[i]); } if(!flag) {work();/*puts("haha");*/return; } int num = 0; for(int i = 1;i <= n;++ i) { a[i] %= p[i],atk[i] %= p[i]; if(!a[i] && !atk[i]) continue; else if(!atk[i]){puts("-1");return;} LL d = gcd(atk[i],p[i]); if(a[i] % d != 0) {continue; } a[i] /= d,atk[i] /= d,p[i] /= d; C[++ num] = mul(a[i] , ((inv(atk[i],p[i]) % p[i] + p[i]) % p[i]),p[i]); M[num] = p[i]; } LL m = M[1],A = C[1],x,y,t; for(int i = 2;i <= num;++ i) { LL d = exgcd(m,M[i],x,y); t = (M[i] / d); //if((C[i]-A)%d == 0 || (a[i] - A) % d == -0 ) { x = mul((x % t + t) % t,(((C[i] - A) / d) % t + t) % t,t); LL MOD = (m / d) * (M[i]); A = (mul(m , x, MOD) + A % MOD) % MOD; m = MOD; // else {puts("-1"); return;}; } A = (A % m + m) % m; printf("%lld\n",A); } int main() { freopen("dragon.in","r",stdin); freopen("dragon.out","w",stdout); int t = read(); while(t --) { init(); } return 0; } 踩
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