hdu4569 Special equations

Special equations

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 184 Accepted Submission(s): 92
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)

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!
   
   
   
   

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

Recommend

zhoujiaqi2010

我们发现题目的数据范围，确定是用int64是可以存下的，我们再看看这题，主要是要求是pri*pri的倍数，那么，我们可以先找到一个是pri的倍数啊，这样，我们就可以通过，这个数不断加上pri，这样就可以大大减少枚举的次数，找到最终的解！如果到了pri*pri,还是没找到解，那么就没有解了，为什么呢？因为，如果一个数大于pri*pri,那么他一定和小于，pri*pri的某一数是相对应的，也就是mod pri*pri 的结果是一样的，因为在pri*pri之内的都不是解，那么大于pri*pri,的自然也没有解了啊！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int a[6],n;
__int64 ff(int x)
{
    int i;
    __int64 sum=0;
    for(i=0;i<n;i++)
    {
        sum=(sum+a[i])*x;
    }
    return sum+a[i];
}
int main()
{
    int tcase,i,j,t=1,pri;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&n);
        for(i=0;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        scanf("%d",&pri);
        printf("Case #%d: ",t++);
        __int64 mod=pri*pri;
        if(n==0&&a[n]%mod!=0)
        {
            printf("No solution!\n");
            continue;
        }
        int temp=0;
        if(ff(0)%pri==0)
        {
            printf("0\n");
            continue;
        }
        for(i=0;i<pri;i++)
        {
            if(ff(i)%pri==0)
            {
                temp=i;
            }

        }
        if(temp==0)
        {
            printf("No solution!\n");
            continue;
        }
        bool flag=true;
        while(temp<mod)
        {
            if(ff(temp)%mod==0)
            {
                printf("%d\n",temp);
                flag=false;
                break;
            }
            temp+=pri;
        }
        if(flag)
            printf("No solution!\n");

    }

    return 0;
}