[BZOJ3751] [NOIP2014] Equation of solution (mathematics)

DescriptionKnown polynomial equation: $a_0+a_1*x+a_2*x^2+...+a_n*x^n=0$ to find the integer solution (n and M are positive integers) within [1,m].InputThe first line contains 2 integers n, m, separated by a space between each of the two integers. The next n+1 line contains an integer for each row, followed by A0,a1,a2,..., an.OutputThe number of integer solutions for the first line of the output equation in [1,m]. Next, an integer in each line, followed by an integer solution of the equation in [1,m], in order from small to large.Sample Input2 10
2
-3
1Sample Output2
1
2HINT

For 100% of data, 0<n≤100,|ai|≤10^10000,an≠0,m≤1000000.

SourceSolution

Considering the equivalent left-side overall confrontation number modulo, this part uses the Qin Jiushao algorithm implementation can greatly speed up, and can avoid a lot of high precision operation

Assuming the modulus is $p$, it is not difficult to find in the $p$ meaning of the $x$ and $x+kp$ obtained the same result, so we can preprocess out $[0,p) $ The answer, the introduction of $[1,m]$ may be the solution

Because $p$ is very small, there will be similar hash collisions happening, so we can select multiple $p$. It is said to choose $5$ a $20000$ around the prime number, but I have changed several prime numbers, not $wa$ is $tle$,qaq

The prime numbers in the code are found on the Internet, and it's unclear why this person's character can be so good qaq,$4$ a prime number $ac$qaq

(The method will be OK, this problem is not the selection of prime numbers, prime numbers copied on the line)

1#include <bits/stdc++.h>2 using namespacestd;3 Chars[ the][10005];4 intp[4] = {17389,22349,22367,22369};5 intN, a[ the], ans[1000005];6 BOOLvis[5][27005];7 8 BOOLCheckintx)9 {Ten     if(!vis[0][x%17389])return false; One     if(!vis[1][x%22349])return false; A     if(!vis[2][x%22367])return false; -     if(!vis[3][x%22369])return false; -     return true; the } -  - BOOLIs_zero (intKintx) - { +     Long LongAns =A[n]; -      for(inti = n-1; ~i; --i) +Ans = (ans * x + a[i])%P[k]; A     return!ans; at } -  - intMain () - { -     intm, tot =0; -scanf"%d%d", &n, &m); in      for(inti =0; I <= N; ++i) -scanf"%s", &s[i]); to      for(inti =0; I <4; ++i) +     { -Memset (A,0,sizeof(a)); the          for(intj =0; J <= N; ++j) *             if(s[j][0] =='-') $                  for(intK =1; S[J][K]; ++k)Panax NotoginsengA[J] = (a[j] *Ten-S[j][k] + -+ p[i])%P[i]; -             Else the                  for(intK =0; S[J][K]; ++k) +A[J] = (a[j] *Ten+ S[j][k]- -) %P[i]; A          for(intj =1; J < P[i]; ++j) theVIS[I][J] =Is_zero (i, j); +     } -      for(inti =1; I <= m; ++i) $         if(check (i)) ans[++tot] =i; $printf"%d\n", tot); -      for(inti =1; I <= tot; ++i) -printf"%d\n", Ans[i]); the     return 0; -}

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[BZOJ3751] [NOIP2014] Equation of solution (mathematics)