[size=x-large]KIDx 的解题报告
[b]该专题必备知识：解模线性方程[/b]
[url]http://972169909-qq-com.iteye.com/blog/1104538[/url]
[color=blue][b]以下原创
转载请指明作者 (KIDx) 以及 文章地址： [/b]
[url]http://972169909-qq-com.iteye.com/blog/1266328[/url][/color]
[/size]
[img]http://dl.iteye.com/upload/attachment/591162/ee09991f-0311-31c0-986f-a838796b7646.png[/img]
[size=medium][color=red][b]问题描述：[/color]给出bi，ni的值，且n1, n2, n3,…, ni两两之间不一定互质，求Res的值？
[color=blue]解：采用的是合并方程的做法。[/color][color=green]
这里将以合并第一第二个方程为例进行说明[/color]
由上图前2个方程得(设k1、k2为某一整数)：[/b][/size]
[img]http://dl.iteye.com/upload/attachment/0068/2935/5e55e37f-5abf-376a-bc59-41f9f93651a2.jpg[/img]
[img]http://dl.iteye.com/upload/attachment/591182/2ff1e981-2405-31ff-87b4-13c2d3715c03.png[/img]
[size=medium]

例题： hdu 1573 X问题 [color=red]【下面已给出代码】[/color]
[url]http://acm.hdu.edu.cn/showproblem.php?pid=1573[/url]
另外推荐一题： hdu 3579 Hello Kiki：
[url]http://acm.hdu.edu.cn/showproblem.php?pid=3579[/url]
[/size]

#include
using namespace std;
#define LL __int64
#define M 10
int N;

LL Egcd (LL a, LL b, LL &x, LL &y)
{
if (b == 0)
{
x = 1, y = 0;
return a;
}
LL d, tp;
d = Egcd (b, a%b, x, y);
tp = x;
x = y;
y = tp - a/b*y;
return d;
}

LL CRT2 (LL b[], LL n[], int num)
{
int i;
bool flag = false;
LL n1 = n[0], n2, b1 = b[0], b2, bb, d, t, k, x, y;
for (i = 1; i < num; i++)
{
n2 = n[i], b2 = b[i];
bb = b2 - b1;
d = Egcd (n1, n2, x, y);
if (bb % d) //模线性解k1时发现无解
{
flag = true;
break;
}
k = bb / d * x; //相当于求上面所说的k1【模线性方程】
t = n2 / d;
if (t < 0) t = -t;
k = (k % t + t) % t; //相当于求上面的K`
b1 = b1 + n1*k;
n1 = n1 / d * n2;
}
if (flag)
return 0; //无解
/******************求正整数解******************/
if (b1 == 0) //如果解为0，而题目要正整数解，显然不行
b1 = n1; //n1刚好为所有ni的最小公倍数，就是解了
/******************求正整数解******************/
if (b1 > N)
return 0;
return (N-b1)/n1+1; //形成的解：b1, b1+n1, b1+2n1,..., b1+xni...
}

int main()
{
int t, num, i, cc = 1;
LL b[M], n[M];
scanf ("%d", &t);
while (t--)
{
scanf ("%d%d", &N, &num);
for (i = 0; i < num; i++)
scanf ("%I64d", n+i);
for (i = 0; i < num; i++)
scanf ("%I64d", b+i);
printf ("%I64d ", CRT2 (b, n, num));
}
return 0;
}