HDU 4569Special equations2012长沙邀请赛E题（数学知识）
Special equations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 178 Accepted Submission(s): 87
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<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!
题目大意：
给你函数 f(x) = a
n
x
n
+...+ a
1
x +a
0 最多N就4位，输入任意一个x使f(x)%(prime*prime)=0。如果找不到就输出NO solution!
解题思路：开始看到中国剩余定理有点懵了，因为以前只是刷过中国剩余定理的一些定理，而且一般都是求很多同余方程。后来有队过了，然后就认真看了下。然后决定暴力做，看了下prime的范围1~10^4,枚举的话就是10^8但是暴力超时了，因为运行测试代码就花了几秒钟。 后来就不解开始看下大白里面讲中国剩余定理。。。再最后半小时左右突然灵光一现，觉得可以先对prime取模得0之后才能继续判断。然后就立马改代码。后来WA了，事实证明当时只是有一个变量写错了。。p+=mo直接每次加mo，怎么会写成了+t！！当时想的也不是很清楚，然后就又换成了p++肯定TLE。 下来之后认真思考了下，发现是没有漏洞的。
思考过程：如果f(x)%(mo*mo)==0的必要条件是f(x)%(mo)==0.然后f(x+k*mo)%mo==0等价于f(x)%mo==0.可以自己拿纸上画一下。当时想的主要是后面有个常数项，然后在想需不需要把它放到右边去TAT！其实是不需要的。 然后就直接在0~mo一个区间里面先找f(x)%mo==0的，如果木有直接输出No sulotion！如果有的话，就+mo判断是否可以
%(mo*mo).....思路主要是根据题目给的说是要把合数分解成几个素数相乘得来的，主要是被中国剩余定理吓到了，需要掌握的知识不能只是了解。加油！
题目地址：Special equations
AC代码：
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
int a[5],n,mo,Mo;
__int64 cal1(int t) //模上mo
{
int i,j;
__int64 res=a[0];
for(i=1;i<=n;i++)
{
__int64 tmp=a[i];
for(j=1;j<=i;j++)
{
tmp=tmp*t;
tmp=tmp%mo;
}
res=(res+tmp)%mo;
}
return res;
}
__int64 cal2(int t) //模上mo*mo
{
int i,j;
__int64 res=a[0];
for(i=1;i<=n;i++)
{
__int64 tmp=a[i];
for(j=1;j<=i;j++)
{
tmp=tmp*t;
tmp=tmp%Mo;
}
res=(res+tmp)%Mo;
}
return res;
}
int main()
{
int tes,cas,t,i,p;
scanf("%d",&tes);
for(cas=1;cas<=tes;cas++)
{
scanf("%d",&n);
int flag=0;
for(i=n;i>=0;i--)
scanf("%d",&a[i]);
scanf("%d",&mo);
Mo=mo*mo;
/*for(i=n;i>=0;i--)
cout<<a[i]<<" ";
cout<<mo<<endl;*/
for(t=0;t<=mo;t++)
{
if(cal1(t)==0) //筛选能模上mo的
{
for(p=t;p<=Mo;p+=mo) //就是这里啊 TAT
{
if(cal2(p)==0)
{
flag=1;
break;
}
}
}
if(flag==1)
break;
}
if(flag==1)
printf("Case #%d: %d\n",cas,p);
else
printf("Case #%d: No solution!\n",cas);
}
return 0;
}