你知道小鱼干藏在哪里了喵

#include #include #include typedef struct stu { int id; char name[9]; int grade;}stu;int comp1(const void*a, const void*b);int comp2(const void*a, const void*b);int comp3(const void*a, co

好久没做树的题啦！#include #include typedef struct TreeNode{ int left; int right;}Node;typedef struct Queue { int data[100]; int front; int rear;}Queue;Node T[20];int cnt = 0,n;Queue Q;void Level

一开始居然怎么都想不到思路。#include #include #include #define MaxSize 100int comp(const void *a, const void *b);int main(){ int N,m,n,i,j,Matrix[MaxSize][MaxSize]; scanf("%d",&N); int *data = (int*)malloc

所以说，乞丐版队列还是很好用的……

#include #include typedef struct Grade { char Name[11]; char ID[11]; char Gender; int grade;}Grade;int main(){ int n,i,FemaleMax=-1,MaleMin=101,FNum,MNum; scanf("%d", &n); Grade *data =

这道题……没想到镜像BST和BST先序怎么转换，直接复制了一下函数。#include #include int IsBST(int Pre[],int begin,int end,int Post[],int PosBegin,int PosEnd);int IsMBST(int Pre[], int begin, int end, int Post[], int PosBegin, in

崩溃啊，第三个测试点格式错误，百度了一下简直拍桌子，没数据的时候要输出两个回车……= =考试的时候遇到这种情况我可能要画圈圈了……控诉一下win10，提交看结果的时候你就断网是不是跟爸爸过不去？？#include #include /*快速排序一趟使主元落在最终位置，因此就是排序好的*/int main(){ int n,i,cnt; scanf("%d", &n); long

把上一题代码稍微改了一下。#include #include typedef struct EdgeNode { int ID; struct EdgeNode* next;}EdgeNode, *PEdgeNode;typedef struct VNode { int UserID; PEdgeNode FirstChild;}VNode;typedef int Posit

用C写真是吃力= =#include #include typedef struct EdgeNode { int UserID; struct EdgeNode* next;}EdgeNode,*PEdgeNode;typedef struct VNode { int UserID; PEdgeNode FirstForward;}VNode;typedef int P

一开始拼命想王道上一道题，好像那是两个序列大小相同的？用了一半的归并。#include #include /*4 1 5 6 83 2 4 8*/int main(){ long n1, n2,i,j,mid,cnt=0,num; scanf("%ld",&n1); long *seq1 = (long*)malloc(sizeof(long)*n1); for (i = 0

这道题以前mooc上做过，记得是差几个测试点，死活AC不了。这次居然一次过了。要注意吃掉判断no之后的数据。思路是要么往pop栈顶元素，要么放入几个之后pop。#include #include int main(){ int m, n, i, k, num,flag,Pop,Top,Max; scanf("%d %d %d", &m, &n, &k); while (k--) {

Merge函数写得乱了，听着鞭炮声敲代码= =#include #include int FindStep(int sorted[], int step,int n);void Merge(int sorted[], int step,int NextStep[],int n);int main(){ int n, i = 0, Isinsertion = 1, Pos; scan

希望都是这种题哈哈。#include #include int comp(const void*a, const void*b);int main(){ int n, i,data[100000],e,j; scanf("%d", &n); for (i = 0; i < n; i++) scanf("%d", &data[i]); qsort(data, n, sizeof(

#include #include typedef struct stu { char name[11]; char id[11]; int grade;}stu;int comp(const void *a, const void*b);int main(){ int n, i,grade1, grade2,flag=0; stu student[1000]; scanf

#include void ConvertPrint(int x);int main(){ int k, card[54], next[54],i,pos,times; scanf("%d", &k); for (i = 0; i < 54; i++) scanf("%d", &next[i]); for (i = 0; i < 54; i++) { pos = i;

#include int main(){ int G[200][200],i,j,VertexNum,EdgeNum,v1,v2,k,SeqSize,IsHC,visited[200]; int seq[1000]; scanf("%d %d", &VertexNum, &EdgeNum); for (i = 0; i < VertexNum; i++) for (j = 0;

#include #define MaxSize 100000typedef struct people { int partner; int IsGuset;}people;int main(){ int n,m,couple1,couple2,guest,cnt,single[MaxSize],i; people data[MaxSize]; for (i = 0; i <

#include #include #include /*能不能不用字符串转化为整数？*/typedef struct Number { long long integer; long long numerator; long long denominator;}Number;long long Convert(char num[], int begin, int end);N

#include typedef struct node { int first; int next;}node;int main(){ node tree[100]; int n, m,i,ChildNum,ID, ParentID,LeavesNum=0; scanf("%d %d", &n, &m); for (i = 0; i <= n; i++) tree[i].

#include typedef struct tree { int first; int next;}tree;int main(){ tree t[100000]; int n, i, supplier, Q[100000], front = 0, rear = 0, x, first = 0, cnt = 1, last = 1,root,level_num=1; doub

L[100000]设为removed的头结点。#include #include typedef struct LinkList { int Key; int Next;}LinkList;int main(){ LinkList L[100001]; int Add, N, FirstAdd,check[10001],i,Removed; scanf("%d %d", &

一开始犯蠢输出直接p*（1+r/100）*（1+r/100）居然还过了两个测试点拿了15分哈哈，机智地骗分~依旧乞丐版队列+乞丐版层次遍历。#include typedef struct tree { int first; int next;}tree;int main(){ tree t[100000]; int n, i, child_num, ID, Q[100000],

#include int FindRoot(int Bird[],int bird);int main(){ int n, i, bird_num,first,bird,root,first_root,bird_sum=0,tree_sum=0; int Bird[10001]; for (i = 1; i <= 10000; i++) Bird[i] = 0; Bird[0]

最近只能在这种简单题上找一找自信了……#include #include int main(){ char sentence[81]; int n1, len,i,j,space; scanf("%s", sentence); len = strlen(sentence); n1 = (len + 2) / 3; for (i = 0,j=len-1; i < n1-1; i+

#include int main(){ int n,i,j,flag,modified=0,m[2000]; char Password[2000][11]; scanf("%d",&n); for (i = 0; i < 2 * n; i++) { scanf("%s", Password[i]); m[i] = 0; } for (i = 1; i < 2 * n;

#include #include int IsPalindromic(int num[],int len);int Add(int num[],int num2[], int len);int main(){ int k,len,i,step=0,num[111],reversed[111]; long long n; scanf("%lld %d", &n,&k); f

#include #define MaxVNum 500#define INF 10000void dj(int l[][MaxVNum],int t[][MaxVNum],int path[], int S,int dist[],int n,int time[]);void dj2(int t[][MaxVNum], int path[], int S, int time[], int

#include #include /*递减减递增的*/int main(){ int num, i=0, x[5], y[5], large,small,ans=0,temp,j; scanf("%d",&num); while (ans != 6174) { for (i = 0; i < 4;i++) { temp = num % 10; num /= 10;

这次写的有点乱= =#include #include int main(){ char Num[10000]; int i, end= 0,exp=0,zero,dot; scanf("%s", Num); while (Num[end] != '\0') end++; end--; i = end; while (Num[i-1]!= 'E') { exp +=(

#include #include /*从两组数中各选出一个相乘，求和的最大值*/void HeapSort(int a[],int n);void HeapAdjust(int a[], int i, int n);int main(){ int ans=0, nc, np,i,j; scanf("%d", &nc); int *Coupon = (int*)malloc(s

两个测试点2、5过不去嘤嘤嘤先放着回头收拾你。#include #include #define MaxSize 17576typedef struct Set { int time; int root; int total_time;}Set;int FindLoc(char p[]);int FindSet(int loc,Set Gang[]);void Convert

当d大于等于MSize时仍没有找到位置，就失败了。但这是为啥？#include #include #include int main(){ int MSize, N,flag,i,x; scanf("%d %d", &MSize, &N); if (MSize == 1) MSize = 2; else if (MSize > 2) { if (MSize % 2==

又一道DFS的题，来自于求树的深度的变体，我是这样解的啦，复习了一下邻接表。其他分析都在注释里了> #include #include /*节点由小到大依次求深度，j指示待插入的位置*//*属于稀疏图，不宜用邻接矩阵*/typedef struct ENode {/*边节点*/ int vetex; struct ENode *next;}ENode,*PENode;typed

不加target==n这个条件测试点1会出现超时=皿=没找到原因。很有价值的贪心题目！#include #define INFINITY 10000/*有时候不用加满油，只要加够向终点的油即可*/typedef struct Station { float price; float dist;}Station;Station GasS[501];int FindTarget(i

T_T心情超级不好啊啊啊啊。这道题想法挺简单的，DFS也做过好多遍了，想到了并查集但没弄清楚怎么做，留个问号在这里吧。选择的是邻接矩阵T口T，所幸有关图的数据量不大。#include /*t偷个懒用邻接矩阵*/int Graph[1000][1000];int visited[1000];int N;void Initvisted();void DFS(int

注意double类型输入为%ld~其他没什么#include /*4，6,6,4 第i个元素，左边i种组合，右边n-i+1种）*/int main(){ int n; double ans=0,num; scanf("%d", &n); for (int i = 1; i <= n;i++) { scanf("%lf",&num); ans += num*i*(n - i

#include #include void HeapAdjust(int a[], int i, int n);/*此处从0开始存放*//*从小到大排用的大根堆，堆顶元素a[0]>a[1]。插入排序a[0]a[1]必定有序，a[0]<a[1]*/int main(){ int n,i=0,Isinsertion=1,Pos; scanf("%d", &n); int *data

两天没刷题= =昨天这道题折腾了我一晚上，排序好后一开始用的递归，测试点3就超时了o(╯□╰)o缩小，。百度到一个很巧妙的方法，从最小数开始试探，满足条件则范围扩大一点，不满足则缩小，一直到范围尾端扩大到最大数，中途记录最大范围。#include#includevoid HeapSort(long a[],int size);void HeapAdjust(long a[], int i

这道题也折磨了好久， 崩溃T_T一开始用的静态链表，二重循环把word1串起来，同样方法串word2，遇到串过的地方就找到了。最后一组大数据超时了。第二次用的链表，最后一组居然内！存！超！过！了！！最后以空间换的时间，反正地址最多十#include typedef struct WordNode { char data; int NextA;}WordNode;WordNode Wor

#include #include #define INFINITY 100000int FindClosest(int a[],int m,int n,int e);/*e表示误差哈哈哈哈哈*//*i表示数列的的始端，若和大于给定值，，应放弃i开头的数列，始端向后移，否则尾端向后移*/int main(){ int n,m, i, e; scanf("%d %d", &n, &