lambsnow的博客

#include <iostream>#include <stdio.h>#include <vector>#include <algorithm>#include <stdlib.h>using namespace std;typedef struct Stu {

#include <stdio.h>#include <stack>#include <deque>using namespace std;int main(int argc, char *argv[]) { int m, n, k, a; scanf("%d %d %d", &m, &n, &k); while (k--) { stack<int> s;

注：如果链表为空，输出 “0 -1”#include <stdio.h>#include <deque>#include <algorithm>#define INC 100000#define MAX 200001using namespace std;int main(int argc, char *argv[]) { int n, start; scanf("%d %

注：比较函数写得有点复杂#include <stdio.h>#include <deque>#include <algorithm>#include <cmath>using namespace std;typedef struct Node { int key; int left; int right;} Node;Node tree[101];int n,

#include <stdio.h>#include <vector>#include <algorithm>#define MAX 1<<26using namespace std;int color[MAX] = {0};bool cmp(int a, int b) { return color[a] > color[b];}int main(int argc, char

注：本以为考察二分法，结果并不是( ´◔ ‸◔’)#include <stdio.h> #include <algorithm>#include <vector>#include <string.h>using namespace std;typedef struct People { char name[9]; int age; int worth;} Peop

#include <stdio.h>#define MAX 9999999typedef struct Ci { int dist; int cost;} Ci;Ci map[501][501];int n, m, s, d;int main(int argc, char *argv[]) { scanf("%d %d %d %d", &n, &m, &s, &d);

#include <stdio.h>#include <string.h>int main(int argc, char *argv[]) { int n, n1, n2, n3; char a[81]; scanf("%s", a); n = strlen(a); if (n % 2 == 0) n2 = 4; else n2 = 3; n

注：vip顾客面临多桌选择时，优先选择vip球桌。注2：非vip顾客有多桌选择时，选id号最小的球桌。注3：只有vip桌空出来的时候，vip顾客才能插队；非vip球桌空出来时，顾客（包括vip与非vip）中到达最早的先选。#include <stdio.h>#include <deque>#include <algorithm>#define OPEN_TIME 8 * 3600#def

#include <stdio.h>#include <iostream>#include <deque>#include <algorithm>#include <string.h>using namespace std;typedef struct Testee { string num; int loca; int score; int rank;

#include <stdio.h>#include <stdio.h>#include <string.h>char in[200];int num[200] = {0};int k;void add(int *a, int &n);bool isPa(int *a, int n);int main(int argc, char *argv[]) { scanf("%s %d

#include <stdio.h>#include <vector>#include <stdlib.h>#include <algorithm>using namespace std;int getId(char *a) { int sum = (a[0]-'A')*26*26*10+(a[1]-'A')*26*10+(a[2]-'A')*10+(a[3]-'0');

#include <stdio.h>int main(int argc, char *argv[]) { int n, m; int face[501] = {0}; //vector<int> v; scanf("%d %d", &n, &m); int i; for (i = 0; i < n; i++) { int tmp;

#include <stdio.h>#include <cmath>int f(int a);int a[11] = {0};int main(int argc, char *argv[]) { int n; scanf("%d", &n); int i, j; int cnt = 0; int tmp = n; while (tmp > 0)

#include <stdio.h>#include <string.h>int main(int argc, char *argv[]) { char a[10001]; char b[10001]; int map[1000] = {0}; gets(a); int al = strlen(a); gets(b); int bl = st

#include <stdio.h>#include <iostream>#include <string.h>#include <queue>#include <stdlib.h>using namespace std;typedef struct cus { int proTime; int serTime; int finTime;} Cus;int n,

注：1是非素数#include <stdio.h>#include <cmath>int getNumWithRadix(int n, int d);int getDecimal(int n, int d); int getReverse(int n);bool isPrime(int n);int main(int argc, char *argv[]) { int n, d;

#include <iostream>#include <iomanip>#include <sstream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <string>#include <map>#include <set>using namespace std;int n;int m;m

#include <stdio.h>#include <string.h>int main(int argc, char *argv[]) { int i, k; char in[23]; int num[23] = {0}; int map[10] = {0}; int map2[10] = {0}; scanf("%s", in); k

#include <iostream>#include <stdio.h>#include <string.h>#include <string>using namespace std;typedef struct per { char ID[16]; int inTime; int outTime;} per;int main(int argc, char *ar

#include <stdio.h>char toChar(int i);int main(int argc, char *argv[]) { int a[3]; scanf("%d %d %d", &a[0], &a[1], &a[2]); int rgb[3][2]; int i; for (i = 0; i < 3; i++) { rgb

此题用Dijkstra计算；如果用Floyd最后一个测试用例会超时#include <stdio.h>#include <iostream> #include <vector>#include <string>#include <algorithm>#include <stdlib.h>#define INF 1<<29using namespace std;struct T {

#include <stdio.h>#include <iostream>#include <stdlib.h>#include <string>using namespace std;int main() { string input; getline(cin, input); int t = input.find('E', 0); string numbe

#include <stdio.h>#include <iostream>using namespace std;struct Node { int e; int next;};Node list[100000];void swap(int *a, int i, int j);int main() { int start, len, k; cin >> sta

注意：如果一名同学提交代码未通过编译，则他这道题成绩打印为0；如果他对某一题从未提交过代码，则他这道题成绩打印为‘-’；如果他提交的所有代码均未通过编译，则不打印他的成绩单。#include <stdio.h>#include <vector>#include <stdlib.h>#include <string.h>#include <algorithm>using namespac

#include <iostream>#include <vector>#include <stack>#include <stdlib.h>#include <string.h>using namespace std;int main() { int n, l; cin >> n >> l; vector<int> *v = new vector<int>[n +

注意：1. getline()可能get到空行，需要过滤一下2. 头文件algorithm中的reverse可以对字符串进行反转#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;int findMinLength(vector<string> v);in

二次探测法：探查序列为d=h(key)，d+1^2，d+2^2，……#include <iostream>using namespace std;bool isPrime(int a);int findPos(int *a, int s, int msize);int main() { int msize, n; cin >> msize >> n; if (!isPr

#include <stdio.h>#include <iostream>#include <vector>#include <stack>#include <math.h>using namespace std;int main() { int n; double p, r; cin >> n >> p >> r; vector<int> *v = new

#include <iostream>#include <string>#include <map>#include <cctype>using namespace std;int main() { string s, tmp; map<string, int> m; getline(cin, s); for (int i = 0; i < s.length(

#include <stdio.h>#include <algorithm>using namespace std;typedef struct Cake { double ton; double price; double up;} Cake;bool cmp(Cake a, Cake b) { return a.up > b.up;}int main(

#include <stdio.h>#include <deque>#include <string.h>#include <algorithm>using namespace std;typedef struct Stu { int id; char name[9]; int grade;} Stu;deque<Stu> student;bool com_i

#include <stdio.h>#include <algorithm>#include <deque>using namespace std;deque<long> dq[2];deque<long> q;int main(int argc, char *argv[]) { for (int i = 0; i < 2; i++) { int m;

注：如果一个用户所有通话记录均不匹配，则不打印他的任何信息#include <stdio.h>#include <string>#include <string.h>#include <vector>#include <algorithm>#include <iostream>using namespace std;typedef struct Rec { string name;

注：这道题除了必要的输入数据外，还可能有杂七杂八的干扰，比如无用的链接、重复的链接等。#include <stdio.h>int main(int argc, char *argv[]) { int list[100000] = {0}; int n; int arr1; int arr2; int set[100000] = {0}; int i;

#include <stdio.h>#include <math.h>#include <stdlib.h>typedef struct Node { Node *left; Node *right; int key;} Node;void insert(Node* &r, int k);int height(Node *r);int max(int a, int

注：第二个for循环中，始终从0开始查找会使测试用例1、2超时，因此作一个小优化，即记住上次开始查找的位置start，下次从此处开始进行查找。#include <stdio.h>void swap(int *a, int *pos, int x, int y) { int tmp = a[x]; a[x] = a[y]; pos[a[y]] = x; a[y] =

#include <stdio.h>#include <cmath>#include <string.h>#include <deque>using namespace std;int main(int argc, char *argv[]) { int np, ng; scanf("%d %d", &np, &ng); int i, j, k, in; in

#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;bool cmp(int a, int b) { return a > b;}int getNum(int *a);int main(int argc, char *argv[]) { int in; int a[

#include <stdio.h>#include <cstdlib>using namespace std;int main(int argc, char *argv[]) { int n; int i, j, k; scanf("%d", &n); int *a = new int[n + 1]; a[0] = 0; for (i = 1; i