"poj2065" SETI

http://poj.org/problem?id=2065 (Topic link)

Test instructions

Test instructions half a day do not understand. Give you a prime number P (p<=30000) and a string of long n strings str[]. The letter ' * ' stands for 0, the letter A-Z represents 1-26, and the numbers represented by the n characters represent F (1), F (2) ... f (n) respectively. Definition: ${f (k) =\sum_{i=0}^{n-1}{a_ik^i~ (mod~p)}~ (1<=k<=n,0<=a_i<p)}$, A0, a1.....an-1. The title guarantee must have the only solution.

Solution

Direct Gaussian elimination, because it is a modular equation group, so in addition to the time to seek an inverse element.

Code

poj2065#include<algorithm> #include <iostream> #include <cstdlib> #include <cstring># include<cstdio> #include <cmath> #include <map> #define LL long long#define inf 2147483640#define Pi ACOs ( -1,0) #define FREE (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout); using namespace Std;const int maxn= 100;int A[maxn][maxn],n,p;char ch[maxn];int Power (int a,int b,int c) {int Res=1;while (b) {if (b&1) res=res*a%c;b>& gt;=1;a=a*a%c;} return res;} void Gauss () {for (int r,i=1;i<=n;i++) {r=i;for (int j=i+1;j<=n;j++) if (ABS (A[r][i]) <abs (A[j][i])) r=j;if (A[r] [i]==0) Continue;if (r!=i) for (int j=1;j<=n+1;j++) swap (a[i][j],a[r][j]), int inv=power (A[I][I],P-2,P); for (int j=1; j<=n;j++) if (j!=i) {for (int k=n+1;k>=i;k--) a[j][k]= (a[j][k]-(A[J][I]*INV)%p*a[i][k]%p+p)%P;}}} int main () {int t;scanf ("%d", &t), while (t--) {scanf ("%d", &p), scanf ("%s", ch+1), N=strlen (ch+1); for (int i=1;i <=n;i++) {int tmp=1;for (int j=1;j<=n;j++) {a[i][j]=tmp;tmp=tmp*i%p;} a[i][n+1]=ch[i]== ' * '? 0:ch[i]-' a ' +1;} Gauss (); for (int i=1;i<=n;i++) {if (a[i][i]==0) {printf ("0"); continue;} int Inv=power (a[i][i],p-2,p);p rintf ("%d", inv*a[i][n+1]%p);}    Puts ("");} return 0;}

　　

"poj2065" SETI