【poj2065】 SETI

http://poj.org/problem?id=2065 (题目链接)

题意

  题意半天看不懂。。给你一个素数P(P<=30000)和一串长为n的字符串str[]。字母'*'代表0,字母a-z分别代表1-26,这n个字符所代表的数字分别代表f(1)、f(2)....f(n)。定义: ${f(k)=\sum_{i=0}^{n-1}{a_ik^i~(mod~p)}~(1<=k<=n,0<=a_i<P)}$,求a0、a1.....an-1。题目保证肯定有唯一解。

Solution

  直接高斯消元,因为是模方程组所以除的时候求个逆元即可。

代码

// 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".in","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>>=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);
			printf("%d ",inv*a[i][n+1]%P);
		}
		puts("");
	}
    return 0;
}

  

posted @ 2016-12-29 16:04  MashiroSky  阅读(...)  评论(...编辑  收藏