hdu3579(中国剩余问题经典)

Description

One day I was shopping in the supermarket. There was a cashier(收银员) counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭，快来快来数一数，二四六七八". And then the cashier put the counted coins back morosely(愁眉苦脸的) and count again...
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.

Input

The first line is T indicating the number of test cases.
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding(相当的) Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi

Output

For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.

Sample Input

14 57

5 56

19 54 40 24 80

11 2 36 20 76

Sample Output

Case 1: 341

Case 2: 5996

题意：res%14=5 res%57=56 求res？

思路：套用中国剩余问题的模板

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<queue>
#include<vector>
using namespace std;
#define N 10005
#define ll __int64
ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
//求一组解(x,y)使得 ax+by = gcd(a,b), 且|x|+|y|最小(注意求出的 x,y 可能为0或负数)。
//下面代码中d = gcd(a,b)
//可以扩展成求等式 ax+by = c,但c必须是d的倍数才有解,即 (c%gcd(a,b))==0
void extend_gcd (ll a , ll b , ll& d, ll &x , ll &y) {  if(!b){d = a; x = 1; y = 0;}else {extend_gcd(b, a%b, d, y, x); y-=x*(a/b);}
}
ll inv(ll a, ll n) { //计算%n下 a的逆。如果不存在逆return -1;ll d, x, y;extend_gcd(a, n, d, x, y);return d == 1 ? (x+n)%n : -1;
}
ll n[N],b[N],len,lcm;
ll work(){for(ll i = 2; i <= len; i++) {ll A = n[1], B = n[i], d, k1, k2, c = b[i]-b[1];extend_gcd(A,B,d,k1,k2);if(c%d)return -1;ll mod = n[i]/d;ll K = ((k1*(b[i]-b[1])/d)%mod+mod)%mod;b[1] = n[1]*K + b[1];n[1] = n[1]*n[i]/d;}if(b[1]==0)return lcm;return b[1];
}
int main(){ll i,T,Cas=1;cin>>T;while(T--){cin>>len;lcm = 1;for(i=1;i<=len;i++) {cin>>n[i];lcm = lcm / gcd(lcm,n[i]) * n[i];}for(i=1;i<=len;i++)cin>>b[i];cout<<"Case "<<Cas++<<": ";cout<<work()<<endl;}return 0;
}

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