Description
Let f(x) = a nx n +...+ a 1x +a 0, in which a i (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 a n to a 0 (0 < abs(a n) <= 100; abs(a i) <= 10000 when deg >= 3, otherwise abs(a i) <= 100000000, i<n). The last integer is prime pri (pri<=10000).
Remember, your task is to solve f(x) 0 (mod pri*pri)
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 a n to a 0 (0 < abs(a n) <= 100; abs(a i) <= 10000 when deg >= 3, otherwise abs(a i) <= 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 many 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!
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题意：给你函数 f(x)最多N就4位，输入任意一个x使f(x)%(prime*prime)=0。
首先要找一个 f(x) % prime = 0 ，只有满足这个条件 f(x) % (prime * prime) 才有可能等于0，对于f(x) % prime，可以枚举 0 - prime，如果满足条件那么在判断（ x + prime * k ) % （prime * prime)是否满足条件
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 typedef long long LL; 7 LL a[5], P; 8 int deg; 9 bool func(LL x, LL mod) 10 { 11 LL res = 0, now = 1; 12 for (int i = 0; i <= deg; i++) 13 { 14 res = (res + now * a[i]) % mod; 15 now = (now * x) % mod; 16 } 17 if (res % mod) 18 return 1; 19 return 0; 20 } 21 int main() 22 { 23 int test; 24 scanf("%d", &test); 25 for (int t = 1; t <= test; t++) 26 { 27 scanf("%d", °); 28 for (int i = deg; i >= 0; i--) 29 { 30 scanf("%I64d", &a[i]); 31 } 32 scanf("%I64d", &P); 33 LL ans = -1; 34 int flag = false; 35 for (LL i = 0; i <= P; i++) 36 { 37 if (func(i, P)) 38 continue; 39 for (LL j = i; j <= P * P; j += P) 40 { 41 if (func(j, P * P)) 42 continue; 43 flag = true; 44 ans = j; 45 break; 46 } 47 if (flag) 48 break; 49 } 50 printf("Case #%d: ", t); 51 if (flag) 52 printf("%I64d ", ans); 53 else 54 printf("No solution! "); 55 } 56 return 0; 57 }