UVa 11040 (water problem) Add bricks in the wall

Test instructions

45 stones are arranged, and the number on each stone is equal to the sum of the two numbers below to support it, and the remaining number is not expressed.

Analysis:

The first is to calculate the number of the bottom line, A71 = A81 + A82 = A91 + 2a92 + A93, the deformation gets A92 = (A71-A91-A93)/2.

And so on, you can get the number of the bottom line. With this "foundation", all the numbers can be counted.

1#include <cstdio>2 3 inta[Ten][Ten];4 5 intMain ()6 {7     //freopen ("In.txt", "R", stdin);8     intT;9scanf"%d", &T);Ten      while(t--) One     { A          for(inti =1; I <=9; i + =2) -              for(intj =1; J <= I; J + =2) -scanf"%d", &a[i][j]); the          for(inti =2; I <=8; i + =2) -a[9][i] = (a[7][i-1]-a[9][i-1]-a[9][i+1])/2; -          for(inti =8; I >=1; --i) -         { +             if((I &1) ==0) -             { +                  for(intj =1; J <= I; ++j) AA[I][J] = a[i+1][J] + a[i+1][j+1]; at             } -             Else -             { -                  for(intj =2; J <= I; J + =2) -A[I][J] = a[i+1][J] + a[i+1][j+1]; -             } in         } -  to          for(inti =1; I <=9; ++i) +         { -              for(intj =1; J < I; ++j) theprintf"%d", A[i][j]); *printf"%d\n", A[i][i]); $         }Panax Notoginseng  -     } the  +     return 0; A}

code June

UVa 11040 (water problem) Add bricks in the wall