解线性同余方程组

poj 21：求方程最小值

//解一元线性同余方程组。

#include <iostream>
#include<cstdio>

using namespace std;

typedef long long ll;

ll ex_(ll a,ll b,ll& x,ll& y)
{
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    ll r,tx,ty;
    r=ex_(b,a%b,tx,ty);
    x=ty;
    y=tx-a/b*ty;
    return r;
}

int main()
{
    ll i,n,a1,r1,a2,r2,ans,a,b,c,d,x0,y0;
    while(scanf("%lld",&n)!=EOF){
        bool flag = 1;
        scanf("%lld%lld",&a1,&r1);
        for( i=1;i<n;i++){
            scanf("%lld%lld",&a2,&r2);
            a = a1;
            b = a2;
            c = r2-r1;
            ll d = ex_(a,b,x0,y0);
            if(c%d!=0){
                flag = 0;
            }
            int t = b/d;
            printf("%lld\n",d);
            printf("%lld\n",x0);
            x0 = (x0*(c/d)%t+t)%t;//保证x0为正
            r1 = a1*x0 + r1;
            a1 = a1*(a2/d);


        }
        if(!flag){
            puts("-1");
            continue;
        }
        printf("%lld\n",r1);
    }
    return 0;
}

hdu1573：求解的大小小于n的解的个数

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll N,m;
ll ex(ll a,ll b,ll &x,ll &y){
	if(b==0){
		x = 1; y = 0; 
		return a;
	}
	ll d = ex(b,a%b,x,y);
	ll tmp = x;
	x = y;
	y = tmp - a/b*x;
	return d;
}
ll solve(ll A[],ll B[],ll num){
	bool ok=true;
	ll a1,b1,a2,b2,k1,kk,d,c,t,x,y;
	a1 = A[0] , b1 = B[0];
	for(ll i=1;i<num;i++){
		a2 = A[i] , b2 = B[i];
		c = b2 - b1;
		d = ex(a1,a2,x,y);
		if( c % d ){	
			ok = false;	break;
		}
		k1 = x*c/d;
		t  = a2/d;
		kk = (k1%t+t)%t;	
		
		b1 = b1 + a1*kk;
		a1 = a1*a2/d;
	}
	if(!ok || b1>N)	return 0;	
	
	ll ans = (N-b1)/a1+1; 
	if(b1==0)	ans--;
	return ans;
}
 
int main(){
	ll T;
	ll A[100],B[100];
	scanf("%lld",&T);
	while(T--){
		scanf("%lld%lld",&N,&m);
		for(int i=0;i<m;i++)	scanf("%lld",&A[i]);
		for(int i=0;i<m;i++)	scanf("%lld",&B[i]);
		ll ans = solve(A,B,m);
		cout<<ans<<endl;
	}
	return 0;
}