hdu4589 Special equations(数论)
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Special equations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 206 Accepted Submission(s): 108
Special Judge
Problem Description
Let f(x) = anxn +...+ a1x +a0, in which ai (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 an to a0 (0 < abs(an) <= 100; abs(ai) <= 10000 when deg >= 3, otherwise abs(ai) <= 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 an to a0 (0 < abs(an) <= 100; abs(ai) <= 10000 when deg >= 3, otherwise abs(ai) <= 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!
Source
2013 ACM-ICPC长沙赛区全国邀请赛——题目重现
题解:f(x)%(p*p)=0那么一定有f(x)%p=0,f(x)%p=0那么一定有f(x+p)%p=0。
所以我们可以开始从0到p枚举x,当f(x)%p=0,然后再从x到p*p枚举,不过每次都是+p,找到了输出即可,没有的话No solution!,这里只需要枚举到pri*pri,这个证法和前面的一样
#include<stdio.h>int a[6],pri,n;__int64 getf(int x,int i){int j;__int64 sum=0;for(j=0;j<i;j++){sum=(sum+a[j])*x;}return sum+a[j];}void solve(){int i,j,k;int pri2=pri*pri;for(i=0;i<pri;i++){if((getf(i,n))%pri==0){for(j=i;j<pri2;j+=pri){if(getf(j,n)%pri2==0){printf("%d\n",j);return ;}}}}printf("No solution!\n");}int main(){int i,j,k,t,no;__int64 temp;scanf("%d",&t);for(k=1;k<=t;k++){scanf("%d",&n);for(i=0;i<=n;i++){scanf("%d",&a[i]);}scanf("%d",&pri);printf("Case #%d: ",k);solve();}return 0;}
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