使用函数判断完全平方数 

没有加（int）过不了 

int IsSquare(int n){if((int)sqrt(n) * sqrt(n) != n) return 0;else return 1;
}

使用函数求余弦函数的近似值 

double funcos(double e, double x){double sum = 1, item = 1;for(int i = 0; fabs(item) >= e; i += 2){item = (-1) * item * x * x / ((i + 1) * (i + 2));sum += item;}return sum;
}

6-13 使用函数输出水仙花数

int narcissistic(int number){int n = number;int item = 0, sum = 0, cnt = 0;while(n){n /= 10;cnt ++;}item = number;while(item){int lastnumber = item % 10;sum += pow(lastnumber, cnt);item /= 10;}if(sum == number) return 1;return 0;}
void PrintN(int m, int n){for(int i = m + 1; i < n; i ++ ){if(narcissistic(i)) printf("%d\n", i);}
}

6-18 使用函数输出指定范围内的完数 

在一些高级语言当中，为了能够完成更好的逻辑判断，因此就有了bool类型，bool类型的变量值只有true和false两种。

而在C语言中，一般认为0为假，非0为真。
这是因为c99之前，c90是没有bool类型的的。但是c99引入了_Bool类型（_Bool就是一个类型，不过在新增头文件stdbool.h中，被重新用宏写成了 bool，为了保证C/C++兼容性）。
目前为止大部分C语言书籍采用的标准还是c90标准，因此我们很少用bool类型。
 

factorsum函数中如果是

int factorsum( int number ){int sum = 1;for(int i = 2; i < number; i ++ ){if(number % i == 0) sum += i;}return sum;
}

过不了 最大范围 的测试点

int factorsum( int number ){int sum = 0;for(int i = 1; i < number; i ++ ){if(number % i == 0) sum += i;}return sum;
}
void PrintPN( int m, int n ){int flag = 0;for(int i = m; i <= n; i ++ ){if(factorsum(i) == i){flag = 1;printf("%d = 1", i);for(int j = 2; j < i; j ++ ){if(i % j == 0) printf(" + %d", j);}printf("\n");}}if(!flag) printf("No perfect number");
}

6-19 使用函数输出指定范围内的Fibonacci数

一开始的代码一直出现 [PTA报错]warning: ignoring return value of ‘scanf’, declared with attribute warn_unused_result 的问题

下面的ac代码反向使用cnt，和我平时的用法感觉妙了点（？

int fib( int n ){if(n == 1 || n == 2) return 1;return fib(n - 1) + fib(n - 2);
}
void PrintFN( int m, int n ){int cnt = 1, flag = 0;for(int i = 1; fib(i) <= n; i ++ ){if(fib(i) >= m && fib(i) <= n){if(cnt){printf("%d", fib(i));cnt = 0;}else printf(" %d", fib(i));flag = 1;}}if(!flag) printf("No Fibonacci number");
}

6-30 删除字符

void delchar( char *str, char c ){int j = 0;for(int i = 0; *(str + i) !='\0'; i ++ ){if(*(str + i) != c){*(str + j) = *(str + i);j ++;}}*(str + j) = '\0';
}

6-31 字符串的连接 

char *str_cat( char *s, char *t ){int len_s = strlen(s);int len_t = strlen(t);for(int i = 0; i < len_t; i ++ ){*(s + len_s + i) = *(t + i);}return &s[0];
}

6-46 指定位置输出字符串 

char *match( char *s, char ch1, char ch2 ){char *p;int i;for(i = 0; i < strlen(s); i ++ ){if(s[i] == ch1) break;}p = &s[i];for(int j = i; j < strlen(s); j ++ ){printf("%c", s[j]);if(s[j] == ch2) break;}printf("\n");return p;
}

6-47 查找子串 

char *search( char *s, char *t ){int flag = 0;for(int i = 0; s[i] != '\0'; i ++ ){for(int j = 0; t[j] != '\0'; j ++ ){flag = 0;if(s[i + j] != t[j]) break;else flag = 1;}if(flag) return &s[i];}return NULL;
}

6-49 建立学生信息链表

void input(){struct stud_node *p;p = (struct stud_node *)malloc(sizeof(struct stud_node));scanf("%d", &p->num);while(p->num != 0){scanf("%s %d", p->name, &p->score);if(head == NULL){head = p;head->next = NULL;}if(tail != NULL) tail->next = p;tail = p;tail->next = NULL;p = (struct stud_node *)malloc(sizeof(struct stud_node));scanf("%d", &p->num);}
}

6-50 学生成绩链表处理 

struct stud_node *createlist(){struct stud_node *head, *tail, *s;head = (struct stud_node *)malloc(sizeof(struct stud_node));head->next = NULL;tail = head;s = (struct stud_node *)malloc(sizeof(struct stud_node));while(1){s=(struct stud_node*)malloc(sizeof(struct stud_node));scanf("%d",&s->num);if(s->num==0)break;scanf("%s",s->name);scanf("%d",&s->score);s->next=NULL;tail->next=s;tail=s;}return head;
}
struct stud_node *deletelist( struct stud_node *head, int min_score ){struct stud_node *p, *n;p = head->next;while(p != NULL){if(p->score < min_score){n = head;while(n->next != p) n = n->next;n->next = p->next;free(p);}p = p->next;}return head->next;
}

6-51 逆序数据建立链表 

struct ListNode *createlist(){struct ListNode *head, *tail, *p;head = tail = NULL;int num;while(scanf("%d", &num) && num != -1){p = (struct ListNode *)malloc(sizeof(struct ListNode));p->data = num;p->next = head;head = p;}return head;
}

6-52 链表拼接 

struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2){struct ListNode *head;head = (struct ListNode *)malloc(sizeof(struct ListNode));struct ListNode *p = head, *p1 = list1, *p2 = list2;while(p1 && p2){if(p1->data < p2->data){p->next = p1;p1 = p1->next;p = p->next;}else{p->next = p2;p2 = p2->next;p = p->next;}}p->next = p1 ? p1 : p2;return head->next;
}

6-53 奇数值结点链表 

struct ListNode *readlist(){struct ListNode *head = NULL;struct ListNode *current, *prev;int n;while(scanf("%d", &n) == 1 && n != -1){current = (struct ListNode *)malloc(sizeof(struct ListNode));if(head == NULL) head = current;else prev->next = current;current->next = NULL;current->data = n;prev = current;}return head;
}
struct ListNode *getodd( struct ListNode **L ){struct ListNode *t = *L;struct ListNode *head_even = NULL, *head_odd = NULL;struct ListNode *current_even, *current_odd, *prev_even, *prev_odd;while(t){if((t->data % 2) == 0){current_even = (struct ListNode *)malloc(sizeof(struct ListNode));if(head_even == NULL) head_even = current_even;else prev_even->next = current_even;current_even->next = NULL;current_even->data = t->data;prev_even = current_even;}else{current_odd = (struct ListNode *)malloc(sizeof(struct ListNode));if(head_odd == NULL) head_odd = current_odd;else prev_odd->next = current_odd;current_odd->next = NULL;current_odd->data = t->data;prev_odd = current_odd;}t = t->next;}*L = head_even;return head_odd;
}

6-54 单链表结点删除 

hhhhh,仿照上面的自己敲出来了红红火火恍恍惚惚哈哈哈（已疯 

struct ListNode *readlist(){struct ListNode *head = NULL;struct ListNode *prev, *current;int n;while(scanf("%d", &n) == 1 && n != -1){current = (struct ListNode *)malloc(sizeof(struct ListNode));if(head == NULL) head = current;else prev->next = current;current->next = NULL;current->data = n;prev = current;}return head;
}
struct ListNode *deletem( struct ListNode *L, int m ){struct ListNode *t = L;struct ListNode *head = NULL;struct ListNode *current, *prev;while(t){if(t->data != m){current = (struct ListNode *)malloc(sizeof(struct ListNode));if(head == NULL) head = current;else prev->next = current;current->next = NULL;current->data = t->data;prev = current;}t = t->next;}return head;
}

6-55 链表逆置 

struct ListNode *reverse( struct ListNode *head ){struct ListNode *Head, *p, *q;p = head;Head = NULL;while(p){q = p->next;p->next = Head;Head = p;p = q;}return Head;
}

6-56 统计专业人数 

int countcs( struct ListNode *head ){int cnt = 0;if(head == NULL) return cnt;while(head){char code[8];strcpy(code, head->code);if(code[1] == '0' && code[2] == '2') cnt ++;head = head->next;}return cnt;
}

 6-57 删除单链表偶数节点

这道题就是前面单链表结点删除直接复制粘贴改几个就ac了（ 

struct ListNode *createlist(){struct ListNode *head = NULL;struct ListNode *prev, *current;int n;while(scanf("%d", &n) == 1 && n != -1){current = (struct ListNode *)malloc(sizeof(struct ListNode));if(head == NULL) head = current;else prev->next = current;current->next = NULL;current->data = n;prev = current;}return head;
}
struct ListNode *deleteeven( struct ListNode *head ){struct ListNode *t = head;struct ListNode *Head = NULL;struct ListNode *current, *prev;while(t){if(t->data % 2 != 0){current = (struct ListNode *)malloc(sizeof(struct ListNode));if(Head == NULL) Head = current;else prev->next = current;current->next = NULL;current->data = t->data;prev = current;}t = t->next;}return Head;}

 7-45 高速公路超速处罚

打%号 %%

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 40;
int main(){int v, lv;cin >> v >> lv;double x = 100.0 * (v - lv) / lv;if(x < 10) cout << "OK";else if(x < 50) printf("Exceed %.lf%%. Ticket 200\n", x);else printf("Exceed %.lf%%. License Revoked\n", x);return 0;
} 

 7-54 猜数字游戏

if else 过不了，要用if if if，因为后面还有一个循环外的条件不能都过？？就是感觉这题还是不太会wwwww

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 40;
int main(){int sn, n;cin >> sn >> n;int cnt = 0;for(int i = 1; i <= n; i ++ ){int a;cin >> a;cnt ++ ;if(a == sn && cnt == 1){cout << "Bingo!" << endl;break;}else if(a == sn && cnt <= 3 && cnt >= 1){cout << "Lucky you!" << endl;break;}else{if(a < 0){cout << "Game over" << endl;break;}if(a > sn){cout << "Too big" << endl;}if(a < sn){cout << "Too small" << endl;}if(a == sn){cout << "Good Guess!" << endl;break;}}if(a != sn && cnt == n){cout << "Game over" << endl;}}return 0;
} 

7-56 高空坠球

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 40;int main(){double H, n;cin >> H >> n;if(n == 0){printf("0.0 0.0");return 0;} double sum = 0, h = H;for(int i = 1; i <= n; i ++ ){sum += h;h /= 2;sum += h;}printf("%.1lf %.1lf", sum - h, h);return 0;
} 

7-57 求分数序列前N项和 

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 40;int main(){int n;cin >> n;double sum  = 0;double z = 2.0, m = 1.0;for(int i = 1; i <= n; i ++ ){sum += z / m;double t = z; z += m;m = t;}printf("%.2lf", sum);return 0;
} 

7-58 求e的近似值 

自定的fun函数int型过不了，要double

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 40;
double fun(int n){double ans = 1;for(int i = 1; i <= n; i ++ ){ans *= i;}return ans;
}
int main(){int n;cin >> n;double sum = 1; for(int i = 1; i <= n; i ++ ){sum += 1.0 / fun(i);}printf("%.8lf", sum);return 0;
} 

7-61 求幂级数展开的部分和 

a啊啊啊啊啊·，还是没做出来，超时了一个点QAQ 

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
//const int N = 40;
int main(){double x;cin >> x;double sum = 1.0, ans = 1.0;for(int i = 1;; i ++ ){ans = (ans * x) / i;sum += ans;if(ans < 1e-5) break;}printf("%.4lf", sum);return 0;
} 

7-71 交换最小值和最大值 

交换位置的时候amax要在amin前面，不然 最小值出现最后 的测试点不能过

分析一下，如果最小值在最后，最大值在最前面的情况下：

eg:8 2 5 4 1

ac的代码来分析，先找到的是和amax相等的a[0]， 交换后就变成了 1 2 5 4 8

继续循环，然后找到了和amin相等的a[0]， 交换后变成了 1 2 5 4 8

拿 最小值出现最后 的测试点不能过的代码来说：

eg:8 2 5 4 1

先找到的是和amax相等的a[0]， 交换后变成了1 2 5 4 8

但是这时候只能继续循环开始 i = 1 了，可是此时最小值在 i = 0 , 所以无法交换最小值到最后的位置 

//最小值出现在最后的测试点过不了
for(int i = 0; i < n; i ++ ){if(amin == a[i]){a[i] = a[0] ^ a[i];a[0] = a[0] ^ a[i];a[i] = a[0] ^ a[i];}if(amax == a[i]){a[n - 1] = a[n - 1] ^ a[i];a[i] = a[n - 1] ^ a[i];a[n - 1] = a[n - 1] ^ a[i];}}

 AC代码

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 10;
int a[N];
int main(){int n;cin >> n;for(int i = 0; i < n; i ++ ) cin >> a[i];int amin = a[0], amax = a[0];for(int i = 1; i < n; i ++ ){amin = min(amin, a[i]);amax = max(amax, a[i]);}for(int i = 0; i < n; i ++ ){if(amax == a[i]){int t = a[n - 1];a[n - 1] = amax;a[i] = t;}if(amin == a[i]){int t = a[0];a[0] = amin;a[i] = t;}}for(int i = 0; i < n; i ++ ) cout << a[i] << " ";return 0;
} 

7-75 找出不是两个数组共有的元素 

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 25;
int a[N], b[N], c[N];
int main(){int n, m, i, j;int k = 0;cin >> n;for(i = 0; i < n; i ++ ) cin >> a[i];cin >> m;for(j = 0; j < m; j ++ ) cin >> b[j];for(i = 0; i < n; i ++ ){for(j = 0; j < m; j ++ ){if(a[i] == b[j]) break;}if(j == m) c[k ++ ] = a[i];}for(i = 0; i < m; i ++ ){for(j = 0; j < n;  j ++ ){if(a[j] == b[i]) break;}if(j == n) c[k ++ ] = b[i];}cout << c[0];for(i = 1; i < k; i ++ ){for(j = 0; j < i; j ++ ){if(c[i] == c[j]) break;}if(j == i) cout << " " << c[i];}return 0;
} 

7-85 螺旋方阵

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 15;
int a[N][N];
int main(){int n;cin >> n;int row = 0, col = n - 1, num = 1;while(row <= col){for(int i = row; i <= col; i ++ ){a[row][i] = num ++ ;}for(int i = row + 1; i <= col; i ++ ){a[i][col] = num ++ ;} for(int i = col - 1; i >= row; i -- ){a[col][i] = num ++ ;}for(int i = col - 1; i >= row + 1; i -- ){a[i][row] = num ++ ;}row ++ , col -- ;}for(int i = 0; i < n; i ++ ){for(int j = 0; j < n; j ++ ){printf("%3d", a[i][j]);}puts("");}return 0;
} 

7-88 统计字符出现次数 

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 350;
int a[N];
int main(){char c;c = getchar();while(c != '\n'){a[c] ++ ;c = getchar();}c = getchar();cout << a[c];return 0;
} 

7-92 字符串转换成十进制整数

 感觉就是有没有做过和看看ASCII的问题

#include 
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
const int N = 1e5 + 10;
LL fun(char *str){LL x = 0;for(int i = 0; i < strlen(str); i ++ ){if(str[i] >= '0' && str[i] <= '9')x = x * 16 + str[i] - '0';else if(str[i] >= 'A' && str[i] <= 'F')x = x * 16 + str[i] - 'A' + 10;}return x;
}
int main(){char s[N], str[N];;cin >> s;int flag = 0, mul = 0, cnt = 0;for(int i = 0; i < strlen(s); i ++ ){s[i] = toupper(s[i]);if(flag == 0){if(s[i] == '-'){flag = 1;mul = 1;}else if((s[i] >= '0' && s[i] <= '9') || (s[i] >= 'A' && s[i] <= 'F')){flag = 1;str[cnt ++ ] = s[i];}}else{if((s[i] >= '0' && s[i] <= '9') || (s[i] >= 'A' && s[i] <= 'F')){str[cnt ++] = s[i];}}}LL ans = fun(str);if(mul == 1) ans *= -1;cout << ans << endl;return 0;
} 

7-103 查找书籍

一开始忘记map还有自动根据key值排序的功能了 

#include 
#define INF 0x3f3f3f3f
#define x first
#define y second
using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
map mp;
int main(){int n;cin >> n;getchar();while(n -- ){string name;double price;getline(cin, name);cin >> price;getchar();mp[price] = name;}cout << fixed << setprecision(2) << (--mp.end())->x << ", " << (--mp.end())->y << endl;cout << fixed << setprecision(2) << mp.begin()->x << ", " << mp.begin()->y << endl;return 0;
} 

7-105 平面向量加法 

一开始nt了一下，进位考虑应该考虑保留最后一位的后一位 

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 1e5 + 10;int main(){double x1, x2, y1, y2;cin >> x1 >> y1 >> x2 >> y2;double x = x1 + x2;double y = y1 + y2;if(fabs(x) < 0.05) x = 0.0;if(fabs(y) < 0.05) y = 0.0;printf("(%.1lf, %.1lf)", x, y);return 0;
} 

 7-108 英文单词排序

哥们觉得vector pair auto 用法需要记一下 

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 1e5 + 10;
typedef pair pr;
vector v;
int cnt;
string t[N];
int main(){string s;while(cin >> s && s != "#"){v.push_back({s.size(), cnt});t[cnt ++ ]  = s;}sort(v.begin(), v.end());for(auto i : v){cout << t[i.y] << " ";}return 0;
} 

7-109 藏头诗

这道题上次做还是int行，改成UTF-8把我干蒙了，触及到只是盲区了

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 1e5 + 10;int main(){string s;for(int i = 0; i < 4; i ++ ){getline(cin, s);cout << s[0] << s[1] << s[2];}return 0;
} 

7-110 自动售货机 

 一开始没注意，被以前的思维带偏了，我为什么要想他一样的物品剩多少匹配多少的问题，NND

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 20;
int a[N];
int sum, total;
int main(){int n;while(cin >> n, n != -1){total += n;}while(cin >> n, n != -1){a[n] ++ ; //编号为n的总数量 if(n > 0 && n < 4) sum ++ ;else if(n > 3 && n < 6) sum += 2;else if(n > 5 && n < 9) sum += 3;else sum += 4;}if(total < sum) cout << "Insufficient money" << endl;else{printf("Total:%dyuan,change:%dyuan\n", total, total - sum);for(int i = 0; i < 11; i ++ ){if(a[i]){switch(i){case 1:printf("Table-water:%d;", a[i]);break;case 2:printf("Table-water:%d;", a[i]);break;case 3:printf("Table-water:%d;", a[i]);break;case 4:printf("Coca-Cola:%d;", a[i]);break;case 5:printf("Milk:%d;", a[i]);break;case 6:printf("Beer:%d;", a[i]);break;case 7:printf("Orange-Juice:%d;", a[i]);break;case 8:printf("Sprite:%d;", a[i]);break;case 9:printf("Oolong-Tea:%d;", a[i]);break;case 10:printf("Green-Tea:%d;", a[i]);break;}}} }return 0;
} 

7-111 停车场管理 

怪，这题真的怪，代码好像可以卡bug 

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 11;
int place[N], tim[N], wait[N];
int waiti = -1;
int main(){int n;cin >> n;while(1){char x;int carnum, cartime;cin >> x >> carnum >> cartime;if(x == 'E') break; else if(x == 'A'){int flag = 0;for(int i = 1; i <= n; i ++ ){if(place[i] == 0){flag = i;break;}}if(flag){place[flag] = carnum;cout << "car#" << carnum << " in parking space #" << flag << endl;}else{wait[++ waiti] = carnum;cout << "car#" << carnum << " waiting" << endl;}}else if(x == 'D'){int flag = 0;for(int i = 0; i <= n; i ++ ){if(place[i] == carnum){flag = i;break;} }if(!flag) cout << "the car not in park" << endl;else{cout << "car#" << carnum << " out,parking time " << tim[flag] << endl;place[flag] = 0;tim[flag] = 0;int t = 0;for(int i = flag; i <= n; i ++ ){place[i] = place[i + 1];tim[i] = tim[i + 1];if(i == n){place[i] = 0;tim[i] = 0;}t = i;}if(waiti != -1){place[t] = wait[0];for(int i = 0; i < waiti; i ++ ){wait[i] = wait[i + 1];}waiti -- ;cout << "car#" << place[t] << " in parking space #" << t << endl;}}}for(int i = 1; i <= n; i ++ ){if(place[i]) tim[i] ++ ;}}return 0;
} 

7-112 值班安排 

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 20;
struct st{int obj1;int obj2;int num;int ret;
}st[N];
int judge(int *d, int n){int ret = 1;for(int i = 0; i < n; i ++ ){if(st[i].ret){if(d[st[i].obj1] != st[i].num - 1){ret = 0;break;}}else if(d[st[i].obj1] + st[i].num != d[st[i].obj2]){ret = 0;break;}}return ret;
}
int main(){int n;cin >> n;getchar();string s;for(int i = 0; i < n; i ++ ){getline(cin, s);st[i].obj1 = s[0] - 'A';if(s[1] == '='){st[i].ret = 1;st[i].num = s[2] - '0';}else{st[i].ret = 0;st[i].obj2 = s[2] - 'A';st[i].num = s[3] - '0';if(s[1] == '<') st[i].num *= -1;}}int day[7];int a, b, c, d, e, f, g;for(a = 0; a < 7; a ++ ){for(b = 0; b < 7; b ++ ){for(c = 0; c < 7; c ++ ){for(d = 0; d < 7; d ++ ){for(e = 0; e < 7; e ++ ){for(f = 0; f < 7; f ++ ){for(g = 0; g < 7; g ++ ){day[0] = a;day[1] = b;day[2] = c;day[3] = d;day[4] = e;day[5] = f;day[6] = g;if(judge(day, n)) goto E;}}}}}}}
E:if(judge(day, n)){char Day[7];for(int i = 0; i < 7; i ++ ){Day[day[i]] = i + 'A';}Day[7] = '\0';puts(Day);}return 0;
} 

7-113 完美的代价

Impossible有两个条件：

1.当n为奇数，如果有两个及以上不能配对的，说明不符合回文串条件

2.当n为偶数，如果相同数字出现奇数次 

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 11;
int main(){int n;string s;cin >> n >> s;int ed = n - 1;int step = 0;bool flag = false;for(int i = 0; i < ed; i ++ ){for(int j = ed; j >= i; j -- ){if(i == j){if(n % 2 == 0 || flag == true){cout << "Impossible"  << endl;return 0;}flag = true;step += n / 2 - i;//当前位置到中间位置需要的步数}else if(s[i] == s[j]){for(int k = j; k < ed; k ++ ){swap(s[k], s[k + 1]);step ++ ;}ed -- ;break;//重点}}}cout << step << endl;return 0;
} 

L2-1 盲盒包装流水线 

一开始只有22分，卡在了开的不够大，紫砂 ，现在st开的数据我不知道具体怎么算的 QAQ，感觉·就看样例的话就是N=可能的最大数据 / S本题样例5 就不会溢出了（

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 1e5 + 10;
stack st[50010];
queue q;
int a[N];
int main(){int n, s, k;cin >> n >> s;for(int i = 1; i <= n; i ++ ){int x;cin >> x;q.push(x);}int col = n / s;for(int i = 0; i < col; i ++ ){for(int j = 0; j < s; j ++ ){int x;cin >> x;st[i].push(x);}for(int j = 0; j < s; j ++ ){int x = q.front();q.pop();int y = st[i].top();st[i].pop();a[x] = y;}}cin >> k;while(k -- ){int x;cin >> x;if(a[x] == 0) cout << "Wrong Number" << endl;else cout << a[x] << endl;}return 0;
} 

L2-2 点赞狂魔 

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 110;
struct person{string name;set tag;double ave;
}p[N];
bool cmp(person x, person y){if(x.tag.size() == y.tag.size()) return x.ave < y.ave;return x.tag.size() > y.tag.size();
}
int main(){int n;cin >> n;for(int i = 0; i < n; i ++ ){string s;cin >> s;p[i].name = s;int k;cin >> k;for(int j = 0; j < k; j ++ ){//woc用while循环过不了LL x;cin >> x;p[i].tag.insert(x);}p[i].ave = 1.0 * k / p[i].tag.size();}sort(p, p + n, cmp);if(n == 0) cout << "- - -" << endl;else if(n == 1) cout << p[0].name << " - -" << endl;else if(n == 2) cout << p[0].name << " " << p[1].name << " -" << endl;else cout << p[0].name << " " << p[1].name << " " << p[2].name << endl;return 0;
} 

L2-3 浪漫侧影

#include 
using namespace std;
#define INF 0x3f3f3f3f
#define x first
#define y second
typedef long long LL;
const int N = 1e5 + 10;
int midt[25], postt[25], tree[N], level;
void createTree(int *postt, int *midt, int n, int idx){if(!n) return ;int k = 0;while(postt[n - 1] != midt[k]) k ++ ;tree[idx] = midt[k];level = max(level, idx);createTree(postt, midt, k, idx * 2);createTree(postt + k, midt + k + 1, n - k - 1, idx * 2 + 1);
}
int main(){int n;cin >> n;for(int i = 0; i < n; i ++ ) cin >> midt[i];for(int i = 0; i < n; i ++ ) cin >> postt[i];createTree(postt, midt, n, 1);for(int i = 1; i <= 22; i ++ ){if(level < pow(2, i)){level = i;break;}}cout << "R: ";int cnt = 0, idx = 1;while(cnt < level){if(tree[idx]) cout << tree[idx];if(tree[idx * 2 + 1]) idx = idx * 2 + 1;else if(tree[idx * 2]) idx *= 2;else{idx *= 2;while(!tree[idx]) idx -- ;}cnt ++ ;if(cnt != level) cout << ' ';else cout << endl;}cout << "L: ";cnt = 0, idx = 1;while(cnt < level){if(tree[idx]) cout << tree[idx];if(tree[idx * 2]) idx *= 2;else if(tree[idx * 2 + 1]) idx = idx * 2 + 1;else{idx *= 2;while(!tree[idx]) idx ++ ;}cnt ++ ;if(cnt != level) cout << ' ';else cout << endl;}return 0;
}