蔡徐坤怎么你辣呜呜呜

又是一道有问题的题目

#include<bits/stdc++.h> using namespace std; bool isPrime[100000010]; //isPrime[i] == 1表示：i是素数 int Prime[6000010], cnt = 0; //Prime存质数、质数表 void GetPrime(int n)//筛到n { memset(isPrime, 1, sizeof(isPrime)); //以“每个数都是素数”为初始状态，逐个删去 isPrime[1] = 0;//1不是素数 } int main() { int n,ans=0; scanf("%d",&n); GetPrime(n); for(int i=1;i<=n;i++) if(Prime[i]) ans++; printf("%d\n", ans); return 0; ———————————————— 版权声明：本文为CSDN博主「qssssss79」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。 原文链接：https://blog.csdn.net/qssssss79/article/details/126029627

我想问一下，比赛中的这个样例组是怎么回事

BEST SPOT 题目描述 Bessie, always wishing to optimize her life, has realized that she really enjoys visiting F (1 <= F <= P) favorite pastures FiF_iFi of the P(1<=P<=500;1<=Fi<=P)P (1 <= P <= 500; 1 <= F_i <= P)P(1<=P<=500;1<=Fi <=P) total pastures (conveniently numbered 1..P) that compose Farmer John's holdings. Bessie knows that she can navigate the C (1 <= C <= 8,000) bidirectional cowpaths (conveniently numbered 1..C) that connect various pastures to travel to any pasture on the entire farm. Associated with each path P_i is a time T_i (1 <= T_i <= 892) to traverse that path (in either direction) and two path endpoints a_i and b_i (1 <= a_i <= P; 1 <= b_i <= P). Bessie wants to find the number of the best pasture to sleep in so that when she awakes, the average time to travel to any of her F favorite pastures is minimized. By way of example, consider a farm laid out as the map below shows, where *'d pasture numbers are favorites. The bracketed numbers are times to traverse the cowpaths. The following table shows distances for potential 'best place' of pastures 4, 5, 6, 7, 9, 10, 11, and 12: Thus, presuming these choices were the best ones (a program would have to check all of them somehow), the best place to sleep is pasture 10. 约翰拥有 P(1<=P<=500)P(1<=P<=500)P(1<=P<=500) 个牧场.贝茜特别喜欢其中的F个.所有的牧场 由 C(1<C<=8000)C(1 < C<=8000)C(1<C<=8000) 条双向路连接，第 iii 路连接着ai,bia_i,b_iai ,bi ，需要 TiT_iTi (1<=Ti<892)(1<=T_i< 892)(1<=Ti <892) 单 位时间来通过. 作为一只总想优化自己生活方式的奶牛，贝茜喜欢自己某一天醒来，到达所有那F个她喜欢的 牧场的平均需时最小.那她前一天应该睡在哪个牧场呢？请帮助贝茜找到这个最佳牧场. 此可见，牧场10到所有贝茜喜欢的牧场的平均距离最小，为最佳牧场. 输入格式 * Line 1: Three space-separated integers: P, F, and C * Lines 2..F+1: Line i+2 contains a single integer: F_i * Lines F+2..C+F+1: Line i+F+1 describes cowpath i with three space-separated integers: a_i, b_i, and T_i 输出格式 * Line 1: A single line with a single integer that is the best pasture in which to sleep. If more than one pasture is best, choose the smallest one. 样例 #1 样例输入 #1 样例输出 #1 提示 As the problem statement As the problem statement.

根本不会做…… / /// ///// /// / / /// ///// /// / / /// ///// /// / / /// ///// /// / / /// ///// /// / / /// ///// /// / 原题链接-<CP002597.[NOIP2010 普及组] 三国游戏>

rfwrn seqeqefafdsdsuidasyhfdufffo9uaedesfdsua请翻译

题目有问题

回原进度之后发现之前的课根本没听懂（大悲 ps：这是求助帖！！！ps：这是求助帖！！！ ps：这是求助帖！！！

有人玩起床吗？ 加我，一起打！！！

#include<bits/stdc++.h> using namespace std; char s[100],q[100]; //q队列 s接受QQ号 int head=0,tail=0;//队首、队尾 //入队 void push(int x){ tail++; q[tail]=x; //q[tail]=x; } //出队 void pop(){ head; } //取队首 char front(){ return q[head+1]; } //取队尾 int back(){ return q[tail]; } //判断队列是否为空 bool empty(){ return headtail; //return tail-head0; } int main(){ cin>>s; for(int i=0;i<strlen(s);i++){ push(s[i]); } while(!empty()){//当队列不为空时 cout<<front(); pop(); push(front()); pop(); } return 0; }