SGU 106 The Equation 扩展欧几里得应用

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Sol：线性不定方程+不等式求解

证明的去搜下别人的证明就好了。。。数学题。

#include <algorithm>#include <cstdio>#include <iostream>using namespace std;long long extend_gcd(long long a,long long b,long long &x,long long &y){if(a==0&&b==0) return -1;if(b==0){x=1;y=0;return a;}long long d=extend_gcd(b,a%b,y,x);y-=a/b*x;return d;}long long a,b,c,x1,x2,y1,y2,ans;int main(){cin>>a>>b>>c>>x1>>x2>>y1>>y2;c=-c;if(a==0&&b==0){if(c==0)  ans=(x2-x1+1)*(y2-y1+1);}else if(a==0){if(c%b==0&&c/b>=y1&&c/b<=y2)ans=x2-x1+1;}else if(b==0){if(c%a==0&&c/a>=x1&&c/a<=x2)ans=y2-y1+1;}else{long long x=0,y=0;long long d=extend_gcd(a,b,x,y);if(c%d==0){long long X=x*(c/d);long long Y=y*(c/d);long long k1,k2,k3,k4;if(x1<=X||(x1-X)*d%b==0)k1=(x1-X)*d/b;elsek1=(x1-X)*d/b+1;if(x2>=X||(X-x2)*d%b==0)k2=(x2-X)*d/b;elsek2=(x2-X)*d/b-1;if(y1<=Y||(y1-Y)*d%a==0)k3=(Y-y1)*d/a;elsek3=(Y-y1)*d/a-1;if(y2>=Y||(Y-y2)*d%a==0)k4=(Y-y2)*d/a;elsek4=(Y-y2)*d/a+1;if(k1>k2) swap(k1,k2);if(k3>k4) swap(k3,k4);ans=min(k2,k4)-max(k1,k3)+1;}}printf("%I64d\n",ans);return 0;}

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