hdu 5084 HeHe (矩阵乘法 找规律)

HeHe

 

n的矩阵

然后m次查询，每次x行乘以y列给ans


从2个查询开始，x,y进行(i+ans)%n的变换

最后输出所有查询ans的和

n3 n4 n5

n2 n3 n4

n1 n2 n3

1 1 的结果是 n3*n3 + n2*n4 + n1*n5  大家可以看到i+j之和为定值，于是就可以预处理啦

#include <iostream>#include <cstdio>#include <climits>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#define esp 1e-6#define inf 0x0f0f0f0f#define LL long long  using namespace std;int x,y,i,j,n,m,a[3000],b[2010][2010],ans; long long int sum=0;int main() {     while(scanf("%d",&n)!=EOF)     {     memset(b,0,sizeof(b));         for(i=0;i<2*n-1;i++)scanf("%d",&a[i]);         for(i=0;i<=2*n-1;i++)        {             b[i][2*n-2]=a[i]*a[2*n-2];            for(j=2*n-1;j>=0;j--)                b[i][j]=b[i-1][j+1]+a[i]*a[j];        }         scanf("%d",&m);        sum=0;        ans=0;        for(i=0;i<m;i++)         {             scanf("%d%d",&x,&y);            x=(x+ans)%n;            y =(y+ans)%n;            ans=b[2*n-x-2][y]-b[n-2-x][n+y];             sum+=ans;        }        printf("%I64d\n",sum);     }     return 0; }