sgu106.The equation 拓展欧几里得 难度:0
106. The equation
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this equation are satisfy to the following conditions : x1<=x<=x2, y1<=y<=y2. Integer root of this equation is a pair of integer numbers (x,y).
Input
Input contains integer numbers a,b,c,x1,x2,y1,y2 delimited by spaces and line breaks. All numbers are not greater than 108 by absolute value.
Output
Write answer to the output.
Sample Input
1 1 -3 0 4 0 4
Sample Output
4
思路:
1 使用欧几里得构造出一组解使ax+by=gcd(a,b),然后(明显c%gcd!=0无解.)两边同乘以(c/gcd)
2 设k1=a/gcd,k2=b/gcd,(x,y)为原方程一组解,那么((x-n*k1),(y+n*k2))也是解(n为任意数)
3 于是不断寻找满足x1<=x<=x2,y1<=y<=y2的解,计数
4 这道题会爆int
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7ffffff;
long long tx1,tx2,ty1,ty2,a,b,c,tx,ty,minn,maxn;
void limit(long long L,long long R,long long d){//注意取区间端点
if(d<0){L=-L;R=-R;d=-d;swap(R,L);}
minn=max(minn,(long long)ceil((double)L/d));
maxn=min(maxn,(long long)floor((double)R/d));
}
long long extgcd(long long a,long long b,long long &x,long long &y){
long long d=a;
if(b!=0){
d=extgcd(b,a%b,y,x);
y-=(a/b)*x;
}
else {
x=1;y=0;
}
return d;
}
int main(){
while(scanf("%I64d%I64d%I64d",&a,&b,&c)==3){
scanf("%I64d%I64d%I64d%I64d",&tx1,&tx2,&ty1,&ty2);
if(tx1>tx2||ty1>ty2){
puts("0");continue;
}
long long ans=0;
if(a==0&&b==0){
if(c==0)ans=(tx2-tx1+1)*(ty2-ty1+1);
}
else if(a==0&&b){
if(c%b==0&&(-c/b)>=ty1&&(-c/b)<=ty2){
ans=(tx2-tx1+1);
}
}
else if(b==0&&a){
if(c%a==0&&(-c/a)>=tx1&&(-c/a)<=tx2){
ans=(ty2-ty1+1);
}
}
else {
int d=extgcd(a,b,tx,ty);
if((-c)%d==0){
tx=-tx*c/d;
ty=-ty*c/d;
minn=-inf;maxn=inf;
limit(tx1-tx,tx2-tx,b/d);
limit(ty1-ty,ty2-ty,-a/d);
if(minn<=maxn)ans=maxn-minn+1;
}
}
printf("%I64d\n",ans);
}
return 0;
}