题目描述 For some years, quite a lot of work has been put into listening to electromagnetic radio signals received from space, in order to understand what civilizations in distant galaxies might be trying to tell us. One signal source that has been of pa…

题目链接:http://poj.org/problem?id=2065 题意:给出一个字符串S[1,n],字母a-z代表1到26,*代表0.我们用数组C[i]表示S[i]经过该变换得到的数字.给出一个素数p.有n个未知数X[1,n].解方程: 思路:消元时让上一个方程乘以一个数下一个方程乘以一个数使得对应位置的数字相等,直接减去即可.最后的a[i][i]*X[i]%p=a[i][n+1]直接枚举X[i]. char s[N]; int a[N][N],n,p,ans[N]; void Gauss…

题目链接 题意: 输入一个素数p和一个字符串s(只包含小写字母和‘*’),字符串中每个字符对应一个数字,'*'对应0,‘a’对应1,‘b’对应2.... 例如str[] = "abc", 那么说明 n=3, 字符串所对应的数列为1, 2, 3. 题目中定义了一个函数: a0*1^0 + a1*1^1+a2*1^2+........+an-1*1^(n-1) = f(1)(mod p), f(1) = str[0] = a = 1; a0*2^0 + a1*2^1+a2*2^2+....…

看题就知道要使用高斯消元求解! 代码如下: #include<iostream> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> using namespace std; ][],p,ans[]; ]; int pows(int a,int b) { ; while(b){ ) ans=(ans*a)%p; b>>=; a=(a*a)%p;…

题意自己看,反正是裸题... 普通高斯消元全换成模意义下行了 模模模! #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; inline int read(){ ,f=; ;c=getchar();} +c-';c=getchar();} return x*f; }…

题目链接 \(Description\) 求\(A_0,A_1,A_2,\cdots,A_{n-1}\),满足 \[A_0*1^0+A_1*1^1+\ldots+A_{n-1}*1^{n-1}\equiv B[1](mod\ p)\] \[A_0*2^0+A_1*2^1+\ldots+A_{n-1}*2^{n-1}\equiv B[2](mod\ p)\] \[\ldots\ldots\ldots\] \[A_0*n^0+A_1*n^1+\ldots+A_{n-1}*n^{n-1}\equiv…

题意: 给出mod的大小,以及一个不大于70长度的字符串.每个字符代表一个数字,且为矩阵的增广列.系数矩阵如下 1^0 * a0 + 1^1 * a1 + ... + 1^(n-1) * an-1 = f(1) 2^0 * a0 + 2^1 * a1 + ... + 2^(n-1) * an-1   = f(2) ........ n^0 * a0 + n^1 * a1 + ... + n^(n-1) * an-1  = f(n) 快速幂取模下系数矩阵 #include <cstdio> #i…

SETI Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1735   Accepted: 1085 Description For some years, quite a lot of work has been put into listening to electromagnetic radio signals received from space, in order to understand what civi…

题目链接:https://vjudge.net/contest/276374#problem/B 题目大意: 输入一个素数p和一个字符串s(只包含小写字母和‘*’),字符串中每个字符对应一个数字,'*'对应0,‘a’对应1,‘b’对应2...,同时题目定义了一个函数:a0*1^0 + a1*1^1+a2*1^2+........+an-1*1^(n-1) = f(1)(mod p), f(1) = str[0] = a = 1; a0*2^0 + a1*2^1+a2*2^2+........+a…

题目大意: f[k] = ∑a[i]*k^i % p 每一个f[k]的值就是字符串上第 k 个元素映射的值,*代表f[k] = 0 , 字母代表f[k] = str[i]-'a'+1 把每一个k^i求出保存在矩阵中,根据字符串的长度len,那么就可以得到len行的矩阵,利用高斯消元解决这个线性方程组 #include <cstdio> #include <cstring> #include <iostream> using namespace std; ; //a[i]…