Update README.md

README.md

@@ -1,4 +1,4 @@
-二叉树
+################################
 #include <vector>
 #include <string>
 #include <stdlib.h>
@@ -96,4 +96,118 @@ int main()
     printf("\n");
     printf("销毁树bt\n");
     DestroyBTree(bt);
-}
+}
+
+##############################
+#include <stdio.h>
+//求解结果表示
+int num;							//全局变量,存放众数
+int maxcnt = 0;						//全局变量,存放重数
+void split(int a[], int low, int high, int& mid, int& left, int& right)
+//以a[low..high]中间的元素为界限,确定为等于a[mid]元素的左、右位置left和right
+{
+	mid = (low + high) / 2;
+	for (left = low; left <= high; left++)
+		if (a[left] == a[mid])
+			break;
+	for (right = left + 1; right <= high; right++)
+		if (a[right] != a[mid])
+			break;
+	right--;
+}
+void Getmaxcnt(int a[], int low, int high)
+{
+	if (low <= high)					//a[low..high]序列至少有1个元素
+	{
+		int mid, left, right;
+		split(a, low, high, mid, left, right);
+		int cnt = right - left + 1;			//求出a[mid]元素的重数
+		if (cnt > maxcnt)				//找到更大的重数
+		{
+			num = a[mid];
+			maxcnt = cnt;
+		}
+		Getmaxcnt(a, low, left - 1);		//左序列递归处理
+		Getmaxcnt(a, right + 1, high);	//右序列递归处理
+	}
+}
+int main()
+{
+	int a[] = { 1,2,2,2,3,3,5,6,6,6,6 };
+	int n = sizeof(a) / sizeof(a[0]);
+	printf("求解结果\n");
+	printf("  递增序列: ");
+	for (int i = 0; i < n; i++)
+		printf("%d ", a[i]);
+	printf("\n");
+	Getmaxcnt(a, 0, n - 1);
+	printf("  众数: %d, 重数: %d\n", num, maxcnt);
+}
+####################################
+#include <iostream>
+using namespace std;
+ 
+const int N=1010;
+int v[N],w[N],f[N];
+int n,m;
+ 
+int main()
+{
+  scanf("%d%d",&n,&m);
+  
+  for(int i=1;i<=n;i++) scanf("%d%d",&v[i],&w[i]);
+  
+  for(int i=1;i<=n;i++)
+    for(int j=m;j>=v[i];j--)
+    {
+      f[j]=max(f[j],f[j-v[i]]+w[i]);
+    }
+    
+  printf("%d",f[m]);
+  
+  return 0;
+}
+######################################
+#include<stdio.h>
+int Fibon1(int n)
+{
+	if (n == 1 || n == 2)
+	{
+		return 1;
+	}
+	else
+	{
+		return Fibon1(n - 1) + Fibon1(n - 2);
+	}
+}
+int main()
+{
+	int n = 0;
+	int ret = 0;
+	scanf("%d", &n);
+	ret = Fibon1(n);
+	printf("ret=%d", ret);
+	return 0;
+}
+
+int Fibno2(int n)
+{
+	int num1 = 1;
+	int num2 = 1;
+	int tmp = 0;
+	int i = 0;
+	if (n < 3)
+	{
+		return 1;
+	}
+	else
+	{
+		for (i = 0; i>n-3; i++)
+		{
+			tmp = num1 + num2;
+			num1 = num2;
+			num2 = tmp;
+		}
+		return tmp;
+	}
+}