Hello Kiki |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 247 Accepted Submission(s): 107 |
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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... |
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Input The first line is T indicating the number of test cases. |
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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. |
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Sample Input 2 2 14 57 5 56 5 19 54 40 24 80 11 2 36 20 76 |
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Sample Output Case 1: 341 Case 2: 5996 |
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Author digiter (Special Thanks echo) |
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Source 2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU |
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Recommend zhouzeyong |
/*
题意：总共有X个钱，分成Mi分会剩余Ai个，让你求X，如果给出的信息不能求出X输出-1
初步思路：中国剩余定理，以前做过类似的题目，韩信点兵，x=m1*m-1*a1+...+mk*mk-1*ak(mod m)
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
/************************中国剩余定理（不互质模板）*****************************/
void exgcd(ll a,ll b,ll& d,ll& x,ll& y)
{
if(!b){d=a;x=1;y=0;}
else
{
exgcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}
ll gcd(ll a,ll b)
{
if(!b){return a;}
gcd(b,a%b);
}
ll M[55],A[55];
ll China(int r)
{
ll dm,i,a,b,x,y,d;
ll c,c1,c2;
a=M[0];
c1=A[0];
for(i=1; i<r; i++)
{
b=M[i];
c2=A[i];
exgcd(a,b,d,x,y);
c=c2-c1;
if(c%d) return -1;//c一定是d的倍数，如果不是，则，肯定无解
dm=b/d;
x=((x*(c/d))%dm+dm)%dm;//保证x为最小正数//c/dm是余数，系数扩大余数被
c1=a*x+c1;
a=a*dm;
}
if(c1==0)//余数为0，说明M[]是等比数列。且余数都为0
{
c1=1;
for(i=0;i<r;i++)
c1=c1*M[i]/gcd(c1,M[i]);
}
return c1;
}
/************************中国剩余定理（不互质模板）*****************************/
int t,n;
int main(){
//freopen("in.txt","r",stdin);
scanf("%d",&t);
for(int ca=1;ca<=t;ca++){
printf("Case %d: ",ca);
scanf("%d",&n);
for(int i=0;i<n;i++) scanf("%lld",&M[i]);
for(int i=0;i<n;i++) scanf("%lld",&A[i]);
printf("%lld\n",China(n));
}
return 0;
}