Hdu4589 Special equations (number theory)

Special equations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 206 Accepted Submission (s): 108
Special Judge

Problem Description
Let f (x) = anxn +... + a1x + a0, in which ai (0 <= I <= n) are all known integers. we call f (x) 0 (mod m) congruence equation. if m is a composite, we can factor m into powers of primes and solve every such single equation after which we merge them using the Chinese Reminder Theorem. in this problem, you are asked to solve a much simpler version of such equations, with m to be prime's square.
 

Input
The first line is the number of equations T, T <= 50.
Then comes T lines, each line starts with an integer deg (1 <= deg <= 4), meaning that f (x)'s degree is deg. then follows deg integers, representing an to a0 (0 <abs (an) <= 100; abs (ai) <= 10000 when deg> = 3, otherwise abs (ai) <= 100000000, I <n ). the last integer is prime pri (pri <= 10000 ).
Remember, your task is to solve f (x) 0 (mod pri * pri)
 

Output
For each equation f (x) 0 (mod pri * pri), first output the case number, then output anyone of x if there are using x fitting the equation, else output "No solution! "
 

Sample Input
4
2 1 1-5 7
1 5-2995 9929
2 1-96255532 8930 9811
4 14 5458 7754 4946-2210 9601

Sample Output
Case #1: No solution!
Case #2: 599
Case #3: 96255626
Case #4: No solution!

Source
2013 ACM-ICPC Changsha Division national invitational Competition -- reproduction of questions
 
Question: f (x) % (p * p) = 0, f (x) % p = 0, f (x) % p = 0 then f (x + p) % p = 0.

So we can start to enumerate x from 0 to p, when f (x) % p = 0, and then enumerate from x to p * p, but every time it is + p, you can find the output. If not, No solution !, Here, you only need to enumerate to pri * pri, which is the same as the previous one.

 

#include<stdio.h>     int a[6],pri,n;    __int64 getf(int x,int i)  {      int j;      __int64 sum=0;      for(j=0;j<i;j++)      {          sum=(sum+a[j])*x;      }      return sum+a[j];  }    void solve()  {      int i,j,k;      int pri2=pri*pri;      for(i=0;i<pri;i++)      {          if((getf(i,n))%pri==0)          {              for(j=i;j<pri2;j+=pri)              {                  if(getf(j,n)%pri2==0)                  {                      printf("%d\n",j);                      return ;                  }              }          }      }      printf("No solution!\n");  }  int main()  {      int i,j,k,t,no;      __int64 temp;      scanf("%d",&t);      for(k=1;k<=t;k++)      {          scanf("%d",&n);          for(i=0;i<=n;i++)          {              scanf("%d",&a[i]);          }          scanf("%d",&pri);          printf("Case #%d: ",k);          solve();      }      return 0;  }  #include<stdio.h>int a[6],pri,n;__int64 getf(int x,int i){int j;__int64 sum=0;for(j=0;j<i;j++){sum=(sum+a[j])*x;}return sum+a[j];}void solve(){int i,j,k;int pri2=pri*pri;for(i=0;i<pri;i++){if((getf(i,n))%pri==0){for(j=i;j<pri2;j+=pri){if(getf(j,n)%pri2==0){printf("%d\n",j);return ;}}}}printf("No solution!\n");}int main(){int i,j,k,t,no;__int64 temp;scanf("%d",&t);for(k=1;k<=t;k++){scanf("%d",&n);for(i=0;i<=n;i++){scanf("%d",&a[i]);}scanf("%d",&pri);printf("Case #%d: ",k);solve();}return 0;}