【POJ】2891 Strange Way to Express Integers

http://poj.org/problem?id=2891

题意：求最小的$x$使得$x \equiv r_i \pmod{ a_i }$。

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <iostream>

using namespace std;

typedef long long ll;

void ex(ll a, ll b, ll &d, ll &x, ll &y) {

	if(!b) { d=a; x=1; y=0; return; }

	ex(b, a%b, d, y, x); y-=a/b*x;

}

ll m[1000005], a[1000005];

int n;

int main() {

	while(~scanf("%d", &n)) {

		for(int i=0; i<n; ++i) scanf("%lld%lld", &m[i], &a[i]);

		ll mm=m[0], aa=a[0]%m[0], d, x, y; int flag=1;

		for(int i=1; i<n; ++i) {

			ll r=a[i]-aa;

			ex(mm, m[i], d, x, y);

			if(r%d) { puts("-1"); flag=0; break; }

			aa+=((x*(r/d)%m[i]+m[i])%m[i])*mm;

			mm=mm/d*m[i];

			aa%=mm;

		}

		if(flag) printf("%lld\n", aa);

	}

	return 0;

}

　　

由$k_1a_1 + r_1 = x = k_2a_2 + r_2$构造出一个$x_0$，所以方程的解满足$x = x_0+k*lcm(a_1, a_2)$。向后递推即可。