HDU-4569 Special equations （数学）

Special equations

http://acm.hdu.edu.cn/showproblem.php?pid=4569

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Special Judge

Problem Description

　　Let f(x) = a_nxⁿ +...+ a₁x +a₀, 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 a₀ (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

42 1 1 -5 71 5 -2995 99292 1 -96255532 8930 98114 14 5458 7754 4946 -2210 9601

 

Sample Output

Case #1: No solution!Case #2: 599Case #3: 96255626Case #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 <cstdio>#include <algorithm>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<T) {        scanf("%I64d",&n);        for(int i=n;i>=0;--i)            scanf("%I64d",a+i);        scanf("%I64d",&pri);        printf("Case #%d: ",++kase);        x=0;        while(x<pri) {            if(f(x)%pri==0)                break;            ++x;        }        if(x==pri)            printf("No solution!\n");        else {            ppri=pri*pri;            while(x<ppri) {                if(f(x)%ppri==0)                    break;                x+=pri;            }            if(x>=ppri)                printf("No solution!\n");            else                printf("%I64d\n",x);        }    }    return 0;}