Hello Kiki |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 258 Accepted Submission(s): 111 |
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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 |
题意：
求同于方程。上模板。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=100005;
const int inf=0x7fffffff;
typedef long long ll;
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y)//扩展欧几里得
{
if(!b) {d=a;x=1;y=0;}
else{
ex_gcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}
ll ex_crt(ll *m,ll *r,int n)
{
ll M=m[1],R=r[1],x,y,d;
for(int i=2;i<=n;i++){
ex_gcd(M,m[i],d,x,y);
if((r[i]-R)%d) return -1;
x=(r[i]-R)/d*x%(m[i]/d);
R+=x*M;
M=M/d*m[i];
R%=M;
}
return R>0?R:R+M;
}
int main()
{
int t,n;
scanf("%d",&t);
for(int cas=1;cas<=t;cas++){
scanf("%d",&n);
ll m[maxn],r[maxn];//m除数，r余数
for(int i=1;i<=n;i++) scanf("%lld",&m[i]);
for(int i=1;i<=n;i++) scanf("%lld",&r[i]);
printf("Case %d: %I64d\n",cas,ex_crt(m,r,n));
}
return 0;
}