#include <iostream> #include <vector> #include <algorithm> #include <qcoreapplication> int main(int argc, char *argv[]) { qcoreapplication a(argc, argv); int minsize, maxsize, n; std::cin >> minsize >> maxsize; std::cin >> n; std::vector<int> fishsize(n); for (int i = 0; i < n; ++i) { std::cin >> fishsize[i]; } std::sort(fishsize.begin(), fishsize.end()); int safecount = 0; for (int size = minsize; size <= maxsize; ++size) { bool issaf
以下是为您补充的 3 点额外需要注意的逻辑点: 1. 要确保输入的数值是有效的,比如不能是负数或者超出合理范围的值。 2. 考虑处理输入错误的情况,比如用户输入的不是整数。 3. 对于边界情况,比如 `minsize` 等于 `maxsize` 或者 `n` 为 0 时,代码的逻辑要能正确处理。 请问您是希望基于这段代码为您做修改完善,还是重新按照新的需求为您编写代码呢? [2024-12-01 23:56:16 | AI写代码神器 | 123点数解答]
相关提问
- [字节豆包] #include <iostream> #include <vector> #include <algorithm> #include <qcoreapplication> int main(int argc, char *argv[]) { qcoreapplication a(argc, argv); int minsize, maxsize, n; std::cin >> minsize >> maxsize; std::cin >> n; std::vector<int> fishsize(n); for (int i = 0; i < n; ++i) { std::cin >> fishsize[i]; } std::sort(fishsize.begin(), fishsize.end()); int safecount = 0; for (int size = minsize; size <= maxsize; ++size) { bool issaf(123点数解答 | 2024-12-01 23:56:16)202
- [字节豆包] #include<iostream> #include<vector> #include <algorithm> using namespace std; int main() { int n; cin>>n; vector <int> v(n); for (int i=0;i<n;i++) { cin>>v[i]; } auto max = max_element(v.begin(), v.end()); auto min = min_element(v.begin(), v.end()); int mmin=min-v.begin(); int mmax=max-v.begin(); int a1=v[0],a2=v[mmin],a3=v[n-1],a4=v[mmax]; v[0]=a2,v[mmin]=a1,v[n-1]=a4,v[mmax]=a3; for (int i=0;i<n;i++) {cout<<v[i]<<" "; } r(113点数解答 | 2024-11-04 11:28:05)259
- [DeepSeek] #include <iostream> #include <vector> #include <algorithm> int main() { int n; std::cin >> n; std::vector<int> times(n); for (int i = 0; i < n; ++i) { std::cin >> times[i]; } std::sort(times.begin(), times.end()); int total = 0; for (int i = 0; i < 5; ++i) { total += times[i]; } std::cout << total << std::endl; return 0; }(263点数解答 | 2026-01-17 10:23:32)18
- [GPT] // // main.cpp // step1 // // created by ljpc on 2018/8/31. // copyright © 2018年 ljpc. all rights reserved. // #include <iostream> #include <stack> #include <cstring> #include <algorithm> using namespace std; int main(int argc, const char * argv[]) { // 请在这里补充代码,完成本关任务 /********* begin *********/ stack<char> s1; stack<char> s2; char s[200]; fgets(s,200,stdin); int len=strlen(s); if(s[len-1]=='\n'){ s[len-1]='\0'; len--; } char(518点数解答 | 2024-10-25 17:28:04)219
- [字节豆包] #include <iostream> #include <sstream> #include <string> #include <cctype> #include <vector> // 将字符串转换为小写 std::string toLowerCase(const std::string& str) { std::string result = str; for (char& c : result) { c = std::tolower(c); } return result; } int main() { std::string target; std::getline(std::cin, target); target = toLowerCase(target); std::string article; std::getline(std::cin, article); std::stringstream ss(article); std::vector<std::(1381点数解答 | 2025-12-15 17:32:28)31
- [字节豆包] #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<vector<int>> A(n, vector<int>(n)); for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { cin >> A[i][j]; } } int maxSum = -1e9; // 计算每行的和 for (int i = 0; i < n; ++i) { int rowSum = 0; for (int j = 0; j < n; ++j) { rowSum += A[i][j]; } maxSum = max(maxSum, rowSum); } // 计算每列的和 for (1151点数解答 | 2025-11-26 20:07:49)25
- [字节豆包] ```cpp #include <iostream> #include <vector> // 定义方块世界的大小 const int WORLD_WIDTH = 10; const int WORLD_HEIGHT = 10; // 定义玩家的位置 int playerX = 0; int playerY = 0; // 定义方块世界 std::vector<std::vector<char>> world(WORLD_HEIGHT, std::vector<char>(WORLD_WIDTH, '.')); // 打印当前的世界状态 void printWorld() { for (int y = 0; y < WORLD_HEIGHT; ++y) { for (int x = 0; x < WORLD_WIDTH; ++x) { if (x == playerX && y == playerY) { std::cout << 'P'; // 玩家位置 } else { (881点数解答 | 2025-12-20 19:47:37)34
- [字节豆包] 采用c++语言,实现如下功能: 实现古典密码中的移位密码和仿射密码,具体实现如下接口: c++接口: <<<<<移位密码>>>>> bool is_valid_s(unsinged char k) { 判断k是否为合理的密钥 } int encrypt_s(unsigned char* p, unsigned char k) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_s(unsigned char* c, unsigned char k) { 密钥合法则返回1,且明文覆盖密文: 密钥不合法则返回0. } <<<<<仿射密码>>>>> bool in_valid_a(unsinged char a, unsigned char b) { 判断a,b是否为合理的密钥 } int encrypt_a(unsigned char* p, unsigned char a, unsigned char b) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_a(unsigned char* c, unsign(812点数解答 | 2024-12-18 16:02:36)331
- [GPT] 运行不了,修改以下代码写出完整代码#include <windows.h> #include <vector> #include <ctime> #include <cstdlib> const int grid_size = 100; // 每个格子的大小 const int box_count = 3; // 箱子数量 const int target_count = 3; // 目标点数量 const int obstacle_count = 5; // 障碍物数量 struct point { int x, y; }; hbitmap hboxbitmap; // 箱子图片的位图句柄 hbrush htargetbrush; // 目标点的画刷 hbrush hobstaclebrush; // 障碍物的画刷 std::vector<point> boxes; // 箱子的位置 std::vector<point> targets; // 目标点的位置 std::vector<point> o(2686点数解答 | 2024-10-31 15:01:39)351
- [字节豆包] 本题要求实现一个函数,查找数组中最大元素的下标。例如数组为{1,3,5,7,9},则该函数应该返回4。题目输入保证数组元素各不相同。 函数接口定义: int indexOfMax(int *array, int size); 其中 array 和 size 是用户传入的参数。 array 是数组首地址; size 是数组元素的个数。函数须返回数组中最大元素的下标(从0开始)。 裁判测试程序样例: /* 此测试程序仅为示例,实际的测试程序可能不同 */ #include <stdio.h> #define MAXSIZE 100 int indexOfMax(int *array, int size); int main(){ int num[MAXSIZE]; int total; scanf("%d",&total); for(int i=0; i<total; i++) scanf("%d",&num[i]); printf("%d\n",num[indexOfMax(num,total)]); return 0; } /* 你(78点数解答 | 2025-12-10 23:10:41)46
- [字节豆包] 0/4 In file included from problem2/problem2.cpp:1:0: problem2/main.h: In function ‘std::__cxx11::string reverseWords(std::__cxx11::string)’: problem2/main.h:8:4: error: ‘vector’ was not declared in this scope vector words; ^~~~~~ problem2/main.h:8:4: note: suggested alternative: ‘perror’ vector words; ^~~~~~ perror problem2/main.h:8:17: error: expected primary-expression before ‘>’ token vector words; ^ problem2/main.h:8:19: error: ‘words’ was not declared in this scope vector words; ^~~~~ probl(489点数解答 | 2025-11-12 14:49:29)54
- [字节豆包] #include<iostream> #include<complex.h> #include <complex> #include <math.h> using namespace std; #define pi 3.14159265358979323846 int main() { file* stream; stream = fopen("4000-red.raw","rb"); unsigned char* image = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); fread(image,4000*4000,sizeof(unsigned char),stream); unsigned char* image2 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); unsigned char* image3 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned(413点数解答 | 2024-11-02 18:16:35)300