Hello Kiki
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2734 Accepted Submission(s): 1010
Problem Description
One
day I was shopping in the supermarket. There was a cashier 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 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.
day I was shopping in the supermarket. There was a cashier 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 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 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
Each
case contains N on the first line, Mi(1 <= i <= N) on the second
line, and corresponding 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.
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
2
14 57
5 56
5
19 54 40 24 80
11 2 36 20 76
Sample Output
Case 1: 341
Case 2: 5996
Case 2: 5996
Author
digiter (Special Thanks echo)
Source
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
using namespace std; /*对于x=r0(mod m0)
x=r1(mod m1)
...
x=rn(mod mn)
输入数组m和数组r，返回[0,[m0,m1,...,mn]-1] 范围内满足以上等式的x0。
x的所有解为:x0+z*[m0,m1,...mn](z为整数)
*/
long long cal_axb(long long a,long long b,long long mod)
{
//防乘法溢出
long long sum=;
while(b)
{
if(b&) sum=(sum+a)%mod;
b>>=;
a=(a+a)%mod;
}
return sum;
} //ax + by = gcd(a,b)
//传入固定值a,b.放回 d=gcd(a,b), x , y
void extendgcd(long long a,long long b,long long &d,long long &x,long long &y)
{
if(b==){d=a;x=;y=;return;}
extendgcd(b,a%b,d,y,x);
y -= x*(a/b);
} long long Multi_ModX(long long m[],long long r[],int n,long long &M)
{
long long m0,r0;
m0=m[]; r0=r[];
for(int i=;i<n;i++)
{
long long m1=m[i],r1=r[i];
long long k0,k1;
long long tmpd;
extendgcd(m0,m1,tmpd,k0,k1);
if( (r1 - r0)%tmpd!= ) return -;
k0 *= (r1-r0)/tmpd;
m1 *= m0/tmpd;
r0 = ( cal_axb(k0,m0,m1)+r0)%m1;
m0=m1;
}
M=m0;
return (r0%m0+m0)%m0;
} int main()
{
int T;
cin>>T;
int tt=;
while(T--)
{
int n;
cin>>n;
long long a[],b[];
for(int i=;i<n;i++)
cin>>a[i];
for(int i=;i<n;i++)
cin>>b[i];
long long M;
long long ans=Multi_ModX(a,b,n,M);
printf("Case %d: ",tt++);
if(ans==) ans+=M;
cout<<ans<<endl;
}
return ;
}