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#include #include #include <stdlib.h> using namespace std; typedef int ElemType; typedef struct node { ElemType data; //数据元素 struct node* lchild; //指向左孩子结点 struct node* rchild; //指向右孩子结点 } BTNode; //二叉链结点类型

BTNode* CreateBTree(ElemType a[], ElemType b[], int n) //对应例2.8的算法由先序序列a[0..n-1]和中序序列b[0..n-1]建立二叉链 { int k; if (n <= 0) return NULL; ElemType root = a[0]; //根结点值 BTNode* bt = (BTNode*)malloc(sizeof(BTNode)); bt->data = root; for (k = 0; k < n; k++) //在b中查找b[k]=root的根结点 if (b[k] == root) break; bt->lchild = CreateBTree(a + 1, b, k); //递归创建左子树 bt->rchild = CreateBTree(a + k + 1, b + k + 1, n - k - 1); //递归创建右子树 return bt; }

void DispBTree(BTNode* bt) //采用括号表示输出二叉链bt { if (bt != NULL) { printf("%d", bt->data); if (bt->lchild != NULL || bt->rchild != NULL) { printf("("); //有孩子结点时才输出( DispBTree(bt->lchild); //递归处理左子树 if (bt->rchild != NULL) printf(","); //有右孩子结点时才输出, DispBTree(bt->rchild); //递归处理右子树 printf(")"); //有孩子结点时才输出) } } }

void DestroyBTree(BTNode*& bt) //释放以bt为根结点的二叉树 { if (bt != NULL) { DestroyBTree(bt->lchild); DestroyBTree(bt->rchild); free(bt); } } int maxsum = 0; //全局变量:存放最大路径和。 vector maxpath; //全局变量:存放最大路径 void Findmaxpath(BTNode* bt, vector apath, int asum) //求根结点到叶结点的路径和最大的路径 { if (bt == NULL) //空树直接返回 return; apath.push_back(bt->data); asum += bt->data; //bt结点加入apath if (bt->lchild == NULL && bt->rchild == NULL) //bt结点为叶结点 { if (asum - maxsum) { maxsum = asum; maxpath.clear(); maxpath = apath; } } Findmaxpath(bt->lchild, apath, asum); //在左子树中查找 Findmaxpath(bt->rchild, apath, asum); //在右子树中查找 } int main() { BTNode* bt; ElemType a[] = { 5,2,3,4,1,6 }; ElemType b[] = { 2,3,5,1,4,6 }; int n = sizeof(a)/sizeof(a[0]); bt = CreateBTree(a,b,n); printf("实验结果:\n"); printf("二叉树bt:"); DispBTree(bt); printf("'in"); printf("最大路径"); vector apath; int asum = 0; Findmaxpath(bt,apath,asum); printf("路径和: % d，路径 : ",maxsum); for (int i=0;i < maxpath.size();i++) printf("%d ",maxpath[i]); printf("\n"); printf("销毁树bt\n"); DestroyBTree(bt); }

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#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 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; } }