主题链接：点击打开链接
特定n*m矩阵，[i,j]分值为gcd(i,j)
给定一个k长的序列，问能否匹配上 矩阵的某一行的连续k个元素
思路：
我们要求出一个解(i,j) 使得 i<=n && j<=m 此时输出 YES
对于j
j % b[0] = 0
j+1 % b[1] = 0
···
j+l % b[l] = 0
依据定理：若 a == b (mod n) => (a+c) == b+c (mod n)
所以将上式变换为
j % b[0] = 0
j % b[1] = -1
···
j % b[n-1] = - (n-1)
最后求出的i,j 检验一下是否满足 gcd(i,j+k) == input[k];
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
#define ll __int64
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
void extend_gcd (ll a , ll b , ll& d, ll &x , ll &y) {
if(!b){d = a; x = 1; y = 0;}
else {extend_gcd(b, a%b, d, y, x); y-=x*(a/b);}
}
ll china(ll l, ll r, ll *m, ll *a){ //下标[l,r] 方程x%m=a;
ll lcm = 1;
for(ll i = l; i <= r; i++)lcm = lcm/gcd(lcm,m[i])*m[i];
for(ll i = l+1; i <= r; i++) {
ll A = m[l], B = m[i], d, x, y, c = a[i]-a[l];
extend_gcd(A,B,d,x,y);
if(c%d)return -1;
ll mod = m[i]/d;
ll K = ((x*c/d)%mod+mod)%mod;
a[l] = m[l]*K + a[l];
m[l] = m[l]*m[i]/d;
}
if(a[l]==0)return lcm;
return a[l];
}
#define N 10005
ll n[N],b[N],tmp[N],len,nn,mm;
bool work(){
memset(b, 0, sizeof b);
memcpy(tmp,n,sizeof n);
ll i = china(1,len,n,b);
if(i>nn || i<=0)return false;
for(ll hehe = 1; hehe <= len; hehe++)b[hehe] = -hehe+1;
memcpy(n,tmp, sizeof n);
ll j = china(1,len,tmp,b);
if(j+len-1>mm || j<=0)return false;
for(ll hehe = 1; hehe <= len; hehe++)if(gcd(i,j+hehe-1)!=n[hehe])return false;
return true;
}
int main(){
while(cin>>nn>>mm>>len){
for(ll i = 1; i <= len; i++) cin>>n[i];
work()? puts("YES"):puts("NO");
}
return 0;
}
版权声明：本文博主原创文章，博客，未经同意不得转载。