HDU-4569 Special equations （数学）
Special equations
http://acm.hdu.edu.cn/showproblem.php?pid=4569
Special Judge
Problem 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 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
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!
看到Special Judge和中国剩余定理就有点怕，直接题解了...发现这对数学好的人来说又是一道简单题
题目只求任意一个x使f(x)%(pri*pri)=0，满足这个的必要条件是f(x)%(pri)=0，所以先在0~pri中枚举x，若找不到符合条件的则无解，找到则令x+=pri再次在x~pri*pri中枚举x，若找不到符合条件的则无解，找到则为答案。
#include
#include
using namespace std;
long long n,a[5],pri,ppri,x;
long long f(long long num) {
long long fx=0,t=1;
for(int i=0;i<=n;++i) {
fx+=a[i]*t;
t*=num;
}
return fx;
}
int main() {
int T,kase=0;
scanf("%d",&T);
while(kase=0;--i)
scanf("%I64d",a+i);
scanf("%I64d",&pri);
printf("Case #%d: ",++kase);
x=0;
while(x=ppri)
printf("No solution!\n");
else
printf("%I64d\n",x);
}
}
return 0;
}