SY_Yu的博客

Problem Description The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified

Problem Description Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.Input Specification: Each input file contai

Problem Description Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places

Problem Description This time, you are supposed to find A+B where A and B are two polynomials.Input Each input file contains one test case. Each case occupies 2 lines, and each line contains the i

Problem Description The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.Input Specification:

Problem Description Given three integers A, B and C in [-263, 263], you are supposed to tell whether A+B > C.Input Specification: The first line of the input gives the positive number of test case

传送门#include<stdio.h>struct P{ char id[20]; int hour,minute,second;}temp,ear,late;//tempbool cmp(struct P a,struct P b){ if(a.hour!=b.hour) return a.hour>=b.hour; if(a.minute!=b.minute

传送门#include<iostream>#include<iomanip>using namespace std;char S[3]={'W','T','L'};int main(){ double num,temp; int max_index; double ans=1; for(int i=0;i<3;i++){ temp=0;

传送门#include<iostream>#include<math.h>using namespace std;struct Student{ char name[20]; char sex; char id[20]; int score;}boy,girl,temp;void init(){ boy.score=101; girl.score

Problem Description Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km.Input Specification: Each input file co

传送门#include<stdio.h>#include<string.h>char str[90];char res[50][50];int main(){ int n,n1,n2; int l; for(int i=0;i<50;i++){ for(int j=0;j<50;j++){ res[i][j]=' ';

Problem Description Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid “inside jobs” where employees col

传送门#include<stdio.h>#include<queue>using namespace std;#define MAX_N 37int inorder[MAX_N],postorder[MAX_N];typedef struct Node{ int data; Node *lchild,*rchild;}node,*Btree;void create(Btree

传送门#include<stdio.h>#include<queue>#include<string.h>#include<stack>using namespace std;#define MAX_N 37int inorder[MAX_N],preorder[MAX_N];typedef struct Node{ int data; Node *lchild,*rchil

传送门#include<stdio.h>#include<queue>#include<string.h>#include<stack>using namespace std;#define MAX_N 37int n;struct Node{ int lchild,rchild;}node[MAX_N];bool visited[MAX_N]={false};int strTonu

传送门#include<stdio.h>#define MAX_N 1000100struct Node{ char data; int next; bool flag;}node[MAX_N];void init(){ for(int i=0;i<MAX_N;i++){ node[i].flag=false; }}int main(){

著名的快速排序算法里有一个经典的划分过程：我们通常采用某种方法取一个元素作为主元，通过交换，把比主元小的元素放到它的左边，比主元大的元素放到它的右边。 给定划分后的N个互不相同的正整数的排列，请问有多少个元素可能是划分前选取的主元？例如给定N = 5, 排列是1、3、2、4、5。则：-1的左边没有元素，右边的元素都比它大，所以它可能是主元； -尽管3的左边元素都比它小，但是它右边的2它小，所以它不

甲级传送门 乙级传送门#include<stdio.h>#include<string.h>#define MAX_N 10010char str[MAX_N];int main(){ gets(str); int l=strlen(str); if(str[0]=='-') printf("-"); int position=0; while(str[p

甲级传送门 乙级传送门#include<stdio.h>#include<math.h>#include <stdlib.h>#include<algorithm>//加上就通过 using namespace std;//加上就通过 typedef long long ll;ll gcd(ll a,ll b){ if(a%b==0) return b; e

传送门#include<iostream>#include<iomanip>using namespace std;struct Poly{ int exp; double coe;}poly[1010];double ans[2020];int main(){ int m,n; int count=0; cin>>m; for(int i=0;

传送门坑点：边界数据0 2输出Yes 0 因此把进制转换部分改成do{ a[i]=m%n; m=m/n; i++;}while(m!=0);更优#include<iostream>using namespace std;int a[100];bool judge(int a[],int i){ for(int j=0;j<=i/2;j++

传送门#include<cstdio>char radix[13]={'0','1','2','3','4','5','6','7','8','9','A' ,'B','C'};int main(){ int m,n,l; scanf("%d%d%d",&m,&n,&l); printf("#"); printf("%c%c",radi

传送门#include<cstdio>int main(){ int A[3],B[3],C[3]; int carry; scanf("%d.%d.%d",&A[0],&A[1],&A[2]); scanf("%d.%d.%d",&B[0],&B[1],&B[2]); C[2]=(A[2]+B[2])%29; carry=(A[2]+B[2])/2

传送门#include<iostream>#include<string>using namespace std;void day(int i){ string s[7]={"MON","TUE","WED","THU","FRI","SAT","SUN"}; cout<<s[i-1]<<" ";}void hour(char c){ if(c<='9'&&c>='0'

甲级传送门乙级传送门#include<stdio.h>#include<algorithm>#define MAX_N 110using namespace std;int n;int aim[MAX_N];int ins[MAX_N];int merg[MAX_N];bool isSame(int a[],int b[]){ bool flag=true; for(int

传送门#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;int N;int m,n;#define MAX_N 11000int a[MAX_N][MAX_N];int num[MAX_N];bool cmp(int a,int b){ return a>b;}void cal_m_n(

问题描述 给定一个正数数列，我们可以从中截取任意的连续的几个数，称为片段。例如，给定数列{0.1, 0.2, 0.3, 0.4}，我们有(0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) (0.4) 这10个片段。 给定正整数数列，

传送门#include<iostream>#include<string>using namespace std;struct people{ string name; string passwd; bool ischange;}p[1010]; void replace(struct people &a,int &cnt){ for(int i=0;i<a.

传送门#include<iostream>#include<vector>using namespace std;vector<string> v;int main(){ int N; string s; int minLen=300; int count=0; cin>>N; //cin.sync();//清空缓冲区 getchar()

甲级传送门 乙级传送门#include<stdio.h>#include<algorithm>using namespace std;#define MAX_N 100100struct Node{ int address; int next; int data; int order;}node[MAX_N];bool cmp(struct Node a,st

传送门#include<stdio.h>#include<algorithm>#include<string.h>#define MAX_N 100100using namespace std;struct Student{ int id; char name[15]; int grade;}stu[MAX_N];bool cmp1(struct Student a,

传送门#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;#define MAX_N 1100struct Record{ char name[25]; int month,d,h,m; bool status;}r[MAX_N],temp;int toll[25];int

甲级传送门 乙级传送门#include<stdio.h>#include<algorithm>#include<string.h> #include<iostream>using namespace std;struct stu{ char id[15]; int de,cai,sum; int flag;}Stu[100100];bool cmp(struct s

传送门#include<stdio.h>#include<string.h>#include<vector>using namespace std;#define MAX_V 510#define INF 1000000000int G[MAX_V][MAX_V];int weight[MAX_V];bool visited[MAX_V];int n,m;int num[MAX_V]

传送门#include<stdio.h>#include<vector>#include<algorithm>#include<string.h>using namespace std;#define MAX_V 510#define INF 100000000int n,m;int Start,End;int G[MAX_V][MAX_V];int d[MAX_V];int cos

传送门#include<stdio.h>#include<algorithm>#include<vector>#define MAX_N 110int ins[MAX_N],heap[MAX_N],aim[MAX_N];int n;void show(int a[]){ for(int i=1;i<=n;i++){ printf("%d",a[i]);

传送门#include<stdio.h>#include<algorithm>using namespace std;#define MAX_N 1100int set[MAX_N];int n;int course[MAX_N];int root[MAX_N];int find(int x){ int r=x; while(set[r]!=r){ r=se

传送门#include<stdio.h>#include<queue>#include<algorithm>#define MAX_N 110using namespace std;struct Node{ int data; int lchild,rchild;}node[MAX_N];int n;int inorder[MAX_N];int co=0;void inO

传送门#include<stdio.h>#include<vector>using namespace std;typedef struct Node{ int data; struct Node *lchild,*rchild; }BTnode,*BTree;int n;vector<int> ori,pre,preM,post,postM;void insert(BTre

传送门#include<stdio.h>#include<algorithm>using namespace std;#define MAX_N 1100int n;int CBSTree[MAX_N]; int in[MAX_N];int co=1;void inOrder(int root){ if(root>n) return; inOrder(root*2);