HDU OJ 3579 Hello Kiki(线性同余方程组的合并求解)

题目：

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2187    Accepted Submission(s): 791

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.

 

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

 

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

2214 575 56519 54 40 24 8011 2 36 20 76

 

Sample Output

Case 1: 341Case 2: 5996

 

Author

digiter (Special Thanks echo)

 

Source

2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU

 

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代码：

/************************************                                  **    @name:HDU OJ 3579             **  @author:npufz                   *************************************/#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>using namespace std;long long gcd(long long a,long long b){    return  (b==0)?a:gcd(b,a%b);}long long exgcd(long long a,long long b,long long &x,long long &y){    if(b==0){x=1;y=0;return a;}    long long  d=exgcd(b,a%b,y,x);    y=y-a/b*x;    return d;}int main(){    long long  m[10],i,t,a[10],cnt,n,t1,x,y;    bool flag;    cin>>t;    for(cnt=1;cnt<=t;cnt++)    {        scanf("%I64d",&n);        for(i=0;i<n;i++)            scanf("%I64d",&m[i]);        for(i=0;i<n;i++)            scanf("%I64d",&a[i]);            flag=true;        for(i=1;i<n&&flag;i++)        {            long long g1;            if(m[0]<m[i])             {                 t1=m[0];m[0]=m[i];m[i]=t1;                 t1=a[0];a[0]=a[i];a[i]=t1;             }            g1=gcd(m[0],m[i]);            long long lcm;            lcm=m[0]*m[i]/g1;            long long c;            c=(a[0]-a[i]);            if(c%g1==0)            {                exgcd(m[0],m[i],x,y);                a[0]=(a[0]-c/g1*x*m[0]);                m[0]=lcm;                a[0]=(a[0]%m[0]+m[0])%m[0];            }            else flag=false;        }        if(flag==false) printf("Case %I64d: %d\n",cnt,-1);        else        {            if(a[0]==0)            printf("Case %I64d: %I64d\n",cnt,m[0]);            else  printf("Case %I64d: %I64d\n",cnt,a[0]);        }}return 0;}

反思：题目是解一组线性同余方程组，用合并同余方程组的方法可以进行求解，这里由于各个除数不是互质的，因此不可以直接套用中国剩余定理，注意如果余数是零，那么输出不可以是零，还有就是用扩展欧几里得解二元一次不定方程的整数解时，注意数字的先后顺序，以免解错，一开始由于在交换两个数据手残大打时错了，就一直以为解方程错了，好久才调试出来，伤死了