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Given four positive integers M, D, L, R, your task is to find the minimal non-negative integer x which satisfies L ≤ ( D × x mod M ) ≤ R, where mod is the remainder operation类欧几里得复杂度log#includecstdio#includecstring#includecstdlib#includealgorithm#includefunctional#includeiostream#includecmath#includecctype#includemap#includectime#includeiomanipusingnamespacestd;#defineFor(i,n)for(inti1;in;i)#defineFork(i,k,n)for(intik;in;i)#defineRep(i,n)for(inti0;in;i)#defineForD(i,n)for(intin;i;i--)#defineRepD(i,n)for(intin;i0;i--)#defineForp(x)for(intppre[x];p;pnext[p])#defineForpiter(x)for(intpiter[x];p;pnext[p])#defineLson(o1)#defineRson((o1)1)#defineMEM(a)memset(a,0,sizeof(a));#defineMEMI(a)memset(a,127,sizeof(a));#defineMEMi(a)memset(a,128,sizeof(a));#defineINF(2139062143)#defineF(100000007)#definepbpush_back#definempmake_pair#definefifirst#definesesecondtypedeflonglongll;llmul(ll a,ll b){return(a*b)%F;}lladd(ll a,ll b){return(ab)%F;}voidupd(lla,ll b){a(a%Fb%F)%F;}mappairpairll,ll,pairll,ll,llh;llcalc(ll l,ll d,ll m,ll r){d%m;rmin(r,m-1);pairpairll,ll,pairll,llphmp(mp(l,d),mp(m,r));// coutl d m rendl;if(lm||lr||r0)return-1;if(!d)return(!l)?0:-1;if(h.count(ph)){// couth[ph]endl;returnh[ph];}h[ph]-1;if(l(ld-1)/d*d(ld-1)/d*dr){returnh[ph](ld-1)/d;}ll ycalc(((-r)%dd)%d,m,d,((-l)%dd)%d);if(y-1)returnh[ph]-1;lm*y,rm*y;returnh[ph](ld-1)/d;}intmain(){// freopen(.in,r,stdin);// freopen(.out,w,stdout);intT;cinT;while(T--){ll m,d,l,r;cinmdlr;coutcalc(l,d,m,r)endl;}return0;}
Given four positive integers M, D, L, R, your task is to find the minimal non-negative integer x which satisfies L ≤ ( D × x mod M ) ≤ R, where mod is the remainder operation类欧几里得复杂度log#includecstdio#includecstring#includecstdlib#includealgorithm#includefunctional#includeiostream#includecmath#includecctype#includemap#includectime#includeiomanipusingnamespacestd;#defineFor(i,n)for(inti1;in;i)#defineFork(i,k,n)for(intik;in;i)#defineRep(i,n)for(inti0;in;i)#defineForD(i,n)for(intin;i;i--)#defineRepD(i,n)for(intin;i0;i--)#defineForp(x)for(intppre[x];p;pnext[p])#defineForpiter(x)for(intpiter[x];p;pnext[p])#defineLson(o1)#defineRson((o1)1)#defineMEM(a)memset(a,0,sizeof(a));#defineMEMI(a)memset(a,127,sizeof(a));#defineMEMi(a)memset(a,128,sizeof(a));#defineINF(2139062143)#defineF(100000007)#definepbpush_back#definempmake_pair#definefifirst#definesesecondtypedeflonglongll;llmul(ll a,ll b){return(a*b)%F;}lladd(ll a,ll b){return(ab)%F;}voidupd(lla,ll b){a(a%Fb%F)%F;}mappairpairll,ll,pairll,ll,llh;llcalc(ll l,ll d,ll m,ll r){d%m;rmin(r,m-1);pairpairll,ll,pairll,llphmp(mp(l,d),mp(m,r));// coutl d m rendl;if(lm||lr||r0)return-1;if(!d)return(!l)?0:-1;if(h.count(ph)){// couth[ph]endl;returnh[ph];}h[ph]-1;if(l(ld-1)/d*d(ld-1)/d*dr){returnh[ph](ld-1)/d;}ll ycalc(((-r)%dd)%d,m,d,((-l)%dd)%d);if(y-1)returnh[ph]-1;lm*y,rm*y;returnh[ph](ld-1)/d;}intmain(){// freopen(.in,r,stdin);// freopen(.out,w,stdout);intT;cinT;while(T--){ll m,d,l,r;cinmdlr;coutcalc(l,d,m,r)endl;}return0;}
