1306 - Solutions to an Equation
| Time Limit: 2 second(s) | Memory Limit: 32 MB |
You have to find the number of solutions of the following equation:
Ax + By + C = 0
Where A, B, C, x, y are integers and x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing seven integers A, B, C, x1, x2, y1, y2 (x1 ≤ x2, y1 ≤ y2). The value of each integer will lie in the range [-108, 108].
Output
For each case, print the case number and the total number of solutions.
Sample Input |
Output for Sample Input |
|
5 1 1 -5 -5 10 2 4 -10 -8 80 -100 100 -90 90 2 3 -4 1 7 0 8 -2 -3 6 -2 5 -10 5 1 8 -32 0 0 1 10 |
Case 1: 3 Case 2: 37 Case 3: 1 Case 4: 2 Case 5: 1 |
PROBLEM SETTER: JANE ALAM JAN
思路：扩展欧几里得；
这题再次让我重新认识了扩欧；其实这题只要分情况讨论下，用扩欧求出特解，然后二分找通解中
t 的范围然后看t重合的部分就行，有几个特殊情况讨论下，还有A，B系数的符号讨论下。
#include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> #include<queue> #include<stack> #include<set> #include<math.h> using namespace std; typedef long long LL; LL gcd(LL n,LL m) { if(m==0) { return n; } else if(n%m==0) { return m; } else return gcd(m,n%m); } pair<LL,LL>P(LL n,LL m) { if(m==0) { pair<LL,LL>ask=make_pair(1,0); return ask; } else { pair<LL,LL>an=P(m,n%m); LL x=an.second; LL y=an.first; y-=(n/m)*x; an.first=x; an.second=y; return an; } } int main(void) { LL i,j,k; scanf("%lld",&k); LL s; LL A,B,C,x1,x2,y1,y2; for(s=1; s<=k; s++) { LL sum=0; scanf("%lld %lld %lld %lld %lld %lld %lld",&A,&B,&C,&x1,&x2,&y1,&y2); C=-C; if(A==0&&B==0&&C!=0) sum=0; else if(A==0&&B==0&&C==0) { sum=(LL)(x2-x1+1)*(LL)(y2-y1+1); } else if(A==0) { if(C%B) { sum=0; } else { LL t=(C/B); if(t>=y1&&t<=y2) sum=(x2-x1+1); else sum=0; } } else if(B==0) { if(C%A) { sum=0; } else { LL t=(C/A); if(t>=x1&&t<=x2) sum=(y2-y1+1); else sum=0; } } else { if(A<0) { C=-C; A=-A; B=-B; } LL gc=gcd(abs(A),abs(B)); if(C%gc) { sum=0; } else if((LL)A*(LL)B>0) { A/=gc; B/=gc; C/=gc; pair<LL,LL>ask=P((A),(B)); LL x=(LL)ask.first; LL y=(LL)ask.second; x*=C; y*=C; LL l=-1e9; LL r=1e9; LL id=1e9; while(l<=r) { LL mid=(l+r)/2; if(x+mid*B>=x1) { id=mid; r=mid-1; } else l=mid+1; } l=-1e9; r=1e9; LL ic=1e9; while(l<=r) { LL mid=(l+r)/2; if(x+mid*B<=x2) { ic=mid; l=mid+1; } else r=mid-1; } if(id>ic) { sum=0; } else if(id==ic) { LL xx=x+id*B; if(xx>=x1&&xx<=x2) { LL yy=y-id*A; if(yy>=y1&&yy<=y2) { sum=1; } } else sum=0; } else { l=-1e9; r=1e9; LL ip=1e9,iq=1e9; while(l<=r) { LL mid=(l+r)/2; if(y-mid*A>=y1) { ip=mid; l=mid+1; } else r=mid-1; } l=-1e9; r=1e9; while(l<=r) { LL mid=(l+r)/2; if(y-mid*A<=y2) { iq=mid; r=mid-1; } else l=mid+1; } if(ip<iq) { sum=0; } else { if(ic<iq||id>ip) { sum=0; } else { if(id<=iq&&ic>=ip) { sum=ip-iq+1; } else if(iq<=id&&ip>=ic) { sum=ic-id+1; } else if(iq>=id&&iq<=ic) { sum=ic-iq+1; } else if(id>=iq&&id<=ip) { sum=ip-id+1; } } } } } else { A/=gc; B/=gc; C/=gc; pair<LL,LL>ask=P(abs(A),abs(B)); LL x=(LL)ask.first; LL y=(LL)ask.second; y=-y; x*=C; y*=C; LL l=-1e9; LL r=1e9; LL id=1e9; while(l<=r) { LL mid=(l+r)/2; if(x+mid*abs(B)>=x1) { id=mid; r=mid-1; } else l=mid+1; } l=-1e9; r=1e9; LL ic=1e9; while(l<=r) { LL mid=(l+r)/2; if(x+mid*abs(B)<=x2) { ic=mid; l=mid+1; } else r=mid-1; } if(id>ic) { sum=0; } else if(id==ic) { LL xx=x+id*abs(B); if(xx>=x1&&xx<=x2) { LL yy=y+id*A; if(yy>=y1&&yy<=y2) { sum=1; } } else sum=0; } else { l=-1e9; r=1e9; LL ip=1e9,iq=1e9; while(l<=r) { LL mid=(l+r)/2; if(y+mid*A>=y1) { iq=mid; r=mid-1; } else l=mid+1; } l=-1e9; r=1e9; while(l<=r) { LL mid=(l+r)/2; if(y+mid*A<=y2) { ip=mid; l=mid+1; } else r=mid-1; } if(ip<iq) { sum=0; } else { if(ic<iq||id>ip) { sum=0; } else { if(id<=iq&&ic>=ip) { sum=ip-iq+1; } else if(iq<=id&&ip>=ic) { sum=ic-id+1; } else if(iq>=id&&iq<=ic) { sum=ic-iq+1; } else if(id>=iq&&id<=ip) { sum=ip-id+1; } } }//printf("%lld %lld %lld %lld ",ip,iq,id,ic); } } } printf("Case %lld: %lld ",s,sum); } return 0; }