Codeforces 598E E. Chocolate Bar (interval dp)

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E. Chocolate Bar

Time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

You had a rectangular chocolate bar consisting of n x m single squares. You want to eat exactly k -squares, so-may need-break the chocolate bar.

In one move can break any single rectangular piece of chocolate in and rectangular pieces. You can break only by lines between squares:horizontally or vertically. The cost of breaking was equal to square of the break length.

For example, if you had a chocolate bar consisting Of 2x3 unit squares then you can Break it horizontally and get Two 1x3 pieces (the cost of such breaking Is 32 = 9), or you can break it vertically in both ways and get, pieces: 2x1 and 2x2  (the cost of such breaking Is 22 = 4).

For several given values n, m and k Find the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set of rectangular Pieces of chocolate with the total size equal to k Squares. The remaining n· M - ksquares is not necessarily form a single rectangular piece.

Input

The first line of the input contains a single integer t (1≤ T ≤40910)-the number of values c4>n, m and K to process.

Each of the next t lines contains three integers n, m and K (1≤ n, m ≤ 30, 1≤ k ≤ min(n· M, ())-the dimensions of the chocolate bar and the number of squares you want to eat respectively.

Output

For each n, m and k Print the minimum total cost needed to break the chocolate bar, In order to make it possible to eat exactly ksquares.

Examplesinput

4
2 2 1
2 2 3
2 2 2
2 2 4

Output

5
5
4
0

Note

In the first query for the sample one needs to perform, breaks:

• To split 2×2 bar into the pieces of 2x1 (cost is 22 = 4),
• To split the resulting 2x1 into a 1x1 pieces (Cost is 12 = 1).

The second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.

AC Code:

#include <bits/stdc++.h>using namespacestd;Const intinf=0x7f7f7f7f;intdp[ +][ +][ -];intn,m,k;intGet_ans () {memset (Dp,inf,sizeof(DP));  for(intI=0; i<= -; i++)    {         for(intj=0; j<= -; j + +) {dp[i][j][0]=0; dp[0][i][j]=0; dp[i][0][j]=0; }    }     for(intI=1; i<= -; i++)    {         for(intj=1; j<= -; j + +)        {            intX=min (I*j, -); if(i*j<= -) dp[i][j][i*j]=0;  for(intk=1; k<=x;k++)            {                 for(intv=1; v<=i/2; v++)                {                    intNum=min (k,v*K);  for(intu=0; u<=num;u++) {Dp[i][j][k]=min (dp[i][j][k],dp[v][j][u]+dp[i-v][j][k-u]+j*K); }                }                 for(intv=1; v<=j/2; v++)                {                    intNum=min (k,v*i);  for(intu=0; u<=num;u++) {Dp[i][j][k]=min (dp[i][j][k],dp[i][v][u]+dp[i][j-v][k-u]+i*H); }                }            }        }    }}intMain () {intT; scanf ("%d",&t);    Get_ans ();  while(t--) {scanf ("%d%d%d",&n,&m,&k); printf ("%d\n", Dp[n][m][k]); }    return 0;}

Codeforces 598E E. Chocolate Bar (interval dp)