1213 number of solutions [one in the theory of numbers with the church practice]

1213 number of solutions

time limit: 1 sspace limit: 128000 KBtitle level: Golden Gold SolvingView Run ResultsTitle Description Description

The known integer x,y satisfies the condition as follows:

Ax+by+c = 0

P<=x<=q

R<=y<=s

The number of x,y that satisfies these conditions.

Enter a description Input Description

The first line has an integer n(n<=10), which indicates that there are n tasks. n<=10

The following are n rows with 7 integers per line:a,b,c,p,Q,R,s. are not more than 108.

Output description Output Description

A total of n rows , and the first row is the number of solutions for the I task.

Sample input Sample Input

2

2 3-7 0 10 0 10

1 1 1-10) 10-9 9

Sample output Sample Output

1

19

Data range and Tips Data Size & HintCategory labels Tags Click here to expandEuclidean theorem number theory Two types of AC code: 1, expand Euclidean algorithm, need to consider special data

#include <iostream>#include<algorithm>#include<math.h>using namespaceStd;typedefLong Longll;Doublez[Ten];voidEXGCD (ll a,ll b,ll &d,ll &x,ll &y) {    if(!B) {d=a;x=1; y=0;return;} EXGCD (B,a%B,D,Y,X); y-=x* (A/b);}intMain () {ll a,b,c,p,q,r,s,x,y,d,k; intN;cin>>N;  while(n--) {cin>>a>>b>>c>>p>>q>>r>>s;c=-C; if(p>q| | r>s| | (!a&&!b&&c)) {cout<<0<<endl;Continue;} if(!a| |!b) {            if(!a) a=q-p+1; Else if(c%a==0&& (C/A) <=q&& (C/A) >=p) a=1; ElseA=0; if(!b) b=s-r+1; Else if(c%b==0&& (c/b) <=s&& (c/b) >=r) b=1; Elseb=0; cout<<a*b<<Endl; Continue;        } exgcd (A,b,d,x,y); if(c%d!=0) {cout<<0<<endl;Continue;} K=c/d;x*=k;y*=K; A=a/d;b=b/D; z[0]= (q-x) *1.0/b; z[1]= (p-x) *1.0/b; z[2]= (y-s) *1.0/A; z[3]= (Y-R) *1.0/A; Sort (z,z+4); A=floor (z[2]); B=ceil (z[1]); cout<<a-b+1<<Endl; }    return 0;}

2. Direct Enumeration

#include <cstdio>#include<iostream>#definell Long Longusing namespacestd;ll N,a,b,c,p,q,r,l,ans;intMain () {CIN>>N;  for(intI=1; i<=n;i++) {cin>>a>>b>>c>>p>>q>>l>>R; Ans=0; if(b==0){             for(intx=p;x<=q;x++){                if(x*a==-c) ans++; } cout<<ans* (r-l+1) <<Endl; }        Else{             for(intx=p;x<=q;x++){                if((-c-a*x)%b==0&& (-c-a*x) *1.0/b>=l&& (-c-a*x) *1.0/B&LT;=R) ans++; } cout<<ans<<Endl; }    }    return 0;}

1213 number of solutions [one in the theory of numbers with the church practice]