转 算法。
1，.N（1<=N<=9）个小熊分一堆苹果，第一只小熊将苹果分成N份，多了一个，扔掉，然后拿走自己的那一份。第二只熊将剩余的苹果分成N份，又多了一个，扔掉，然后拿走自己的那一份，第三只.....，直到第N只熊；问最初的苹果有多少个？
public static int n =6;
static void Main(string[] args)
{
for (int i = 15550; i <= Int32.MaxValue; i++)
{
if (solve(i*3+1))
{
Console.WriteLine(n + " " + (i * 3 + 1));
break;
}
}
}
public static bool solve(int apple)
{
int b1 = (apple - 1) / 3;
for (int i = 2; i < n + 1; i++)
{
if (!(((Math.Pow(n - 1, i - 1)) * (apple + n - 1)) % (Math.Pow(n, i)) == 0))
{ return false; }
}
return true;
}2.6*6的矩阵，从左上方开始，只经过向下或向右的步骤，到达右下方，找出经过的位置的最大价值； 200,120,400,150,180,300 150,250,360,120,200,130 350,300,250,100,500,260 100,150,260,320,100,150 500,130,260,100,200,170 160,100,250,200,600,200
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace test2
{
class Program
{
static void Main(string[] args)
{
int[][] value = new int[][] {new int[]{200,120,400,150,180,300},
new int[] {150,250,360,120,200,130},
new int[] {350,300,250,100,500,260},
new int[] {100,150,260,320,100,150},
new int[] {500,130,260,100,200,170},
new int[] {160,100,250,200,600,200}};
work(value);
//show(value);
}
public static int work(int[][] value){
int[][] maxValue = new int[][]{
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
};
int[][] path = new int[][]{
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
};
maxValue[0][0] = value[0][0];
for (int i = 1; i < 6; i++) {
maxValue[0][i] = value[0][i]+maxValue[0][i-1];
path[0][i] =1;
}
for (int i = 1; i < 6; i++) {
maxValue[i][0] = value[i][0]+maxValue[i-1][0];
path[0][i] = 1;
}
for (int i = 1; i < 6; i++) {
for (int j = 1; j < 6; j++) {
maxValue[i][j] =Math.Max(maxValue[i-1][j], maxValue[i][j-1])+value[i][j];
if (maxValue[i - 1][j] > maxValue[i][j - 1])
{ path[i][j] = 0; }
else { path[i][j] = 1; }
}
}
show(maxValue);
Console.WriteLine();
show(path);
Console.WriteLine();
showstr(path);
return maxValue[5][5];
}
public static void show(int[][] mar)
{
foreach (int[] i in mar)
{
foreach (int x in i)
{
Console.Write(x.ToString() + " ");
}
Console.WriteLine();
}
}
public static void showstr(int[][] mar)
{
int[][] path = new int[][]{
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
new int[]{0,0,0,0,0,0},
};
mar[0][0] = 1; mar[5][5] = 1;
int i = 5; int j = 5;
while (true)
{
if (!(i == 0 & j == 0))
{
if (mar[i][j] == 0)
{
path[--i][j] = 1;
}
else
{
path[i][--j] = 1;
}
}
else { break; }
}
show(path);
}
}
}思路最上，最左向右累加，其他部分取上方左方最大，累加下去。
路径思路，上面向右设为1，向下设为0，在第二个矩阵里进值，路径自最右下开始按方向开始，本题左左左上上上上上左左。结果为矩阵三
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