# 51nod-1346: Recursion

Source: Internet
Author: User

"Portal: 51nod-1346" Brief test Instructions:

Give a formula A[i][j]=a[i-1][j]^a[i][j-1]

Give A[1][i],a[i][1] (2<=i<=131172)

There are n queries, each asking input x, y

Find out a[x+131072][y+131072]

The following:

It's really an interesting topic.

Obviously can't do it directly, and then I hit a table directly. Found A[i][j]=a[i-k][j]^a[i][j-k] (k<=min (I,j), and K is even)

Direct violence get a[1~100][1~131172] and a[1~131172][1~100] on the line

Reference Code:
`#include <cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>using namespacestd;inta[ the][131173];intb[131173][ the];intMain () { for(intI=2; i<=131172; i++) {scanf ("%d", &a[1][i]); if(i<= -) b[1][i]=a[1][i]; }     for(intI=2; i<=131172; i++) {scanf ("%d", &b[i][1]); if(i<= -) a[i][1]=b[i][1]; }     for(intI=2; i<= -; i++)    {         for(intj=2; j<=131172; j + +) a[i][j]=a[i-1][j]^a[i][j-1]; }     for(intI=2; i<=131172; i++)    {         for(intj=2; j<= -; j + +) b[i][j]=b[i-1][j]^b[i][j-1]; }    intN; scanf ("%d",&N);  while(n--)    {        intx, y; scanf ("%d%d",&x,&y); printf ("%d\n", a[x][y+131072]^b[x+131072][y]); }    return 0;}`

51nod-1346: Recursion

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