Description

One day I was shopping in the supermarket. There was a cashier(收銀員)?counting coins seriously when a little kid running and singing "門前大橋下游過一群鴨，快來快來 數一數，二四六七八". And then the cashier put the counted coins back morosely(愁眉苦臉的)?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(相當的)?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

2

2

14 57

5 56

5

19 54 40 24 80

11 2 36 20 76

Sample Output

Case 1: 341

Case 2: 5996

題意：res%14=5 res%57=56 求res？

思路：套用中國剩余問題的模板

代碼：

#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); } //求一組解(x,y)使得 ax+by = gcd(a,b), 且|x|+|y|最小(注意求出的 x,y 可能為0或負數)。 //下面代碼中d = gcd(a,b) //可以擴展成求等式 ax+by = c,但c必須是d的倍數才有解,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) { //計算%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; }

總結

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