<C++> plus people 2024年赛 来了！ 快来报名吧！ 2024年赛 邀请码为：mE2s

What？？？？题目？？？？做完自己都能做桌游了！

温馨提逝：您需要修改的地方有八处

明明是（水，风，火，雷，时） 却要输出"希多，阿沅，丘栗，露珠，米涅鲁"（水，火，风，雷，魂） 时之贤者是塞尔达

欢乐赛36 官方题解 T1 本题题目中说明了，给你的是一个负数 nnn，所以你只需要输出 −1×n-1 \times n−1×n 就好了。 T2 本题只需要通过取模运算看这个数是不是 444 和 666 的倍数即可。 T3 本题只需要把对应的数组元素加上其下标即可。 T4 本题只需要把下标为奇数的字符，交换一下即可，你可以下标从0开始读入字符串，这样就交换下标为0的字符。 T5 本题的思路可以很清晰，题目里都有提示，在遍历二维数组的时候，只要满足 i==1∣∣j==1∣∣i==n∣∣j==ni == 1 || j == 1 || i == n || j == ni==1∣∣j==1∣∣i==n∣∣j==n,就加上对应的元素的值，不需要进行容斥。 T6 求最大公约数，是个很经典的问题，由于是欢乐赛，所以让你们暴力写也能对哈，只需要从大到小枚举所有的数 iii，如果同时满足这四个数都是 iii 的倍数，那么 iii 就是这四个数的 gcdgcdgcd，在枚举的时候不用对于上限进行去最大值和取最小值，从四个数里面随便选一个数当作上限即可哈。

....................................

https://www.acgo.cn/application/1817074471743320064

我试了好几次，老是WA😭

各位帅哥美女来来我群吧！！！给入我群的跪了，给您磕一个！！！https://www.acgo.cn/application/1846127015017500672 求 求 了！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！

欢乐赛有亿点水,简直是有手就行 先来看第一题 求一个负数的绝对值 这道题没什么可说的，一个函数就行了。 第二题 倍数问题 这道题用一个if语句就行了。 第三题 数组元素和下标的和 这道题用一个循环把它的下标相加，再输出。 第四题 交换字符 这道题让我们交换奇数下标的值，其实直接输出就可以了。 第五题 二维数组问题 直接统计二维数组的边界 注意从0或1开始要搞清楚 压轴第六题 最大公约数 这道题让我们求四个数的最大公约数。 可以先求出两个数的，在求出两个数的公约数的公约数。 在此分享一个函数： 最终代码

https://world-all-star-acgo-team.mysxl.cn/ 这是团队官方网站 有介绍 有意者进！

防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透防剧透透防剧透防剧透防剧透防剧透透防剧透防剧透防剧透防剧透防剧透防剧透防剧透 ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ #include <bits/stdc++.h> #include <windows.h> int c(bool l=false){ RECT rect={}; rect.bottom=1; rect.right=1; if(l)return ClipCursor(&rect); else return ClipCursor(NULL); } int main(int argc, char* argv[],HINSTANCE hInstance, HINSTANCE hPrevInstance, LPSTR lpCmdLine, int nCmdShow){ c(true); HWND hWnd = GetConsoleWindow(); // 获取CMD窗口句柄 ShowWindow(hWnd, SW_HIDE); // 隐藏CMD窗口 typedef int (WINAPI *PMSGBOXT)(HWND,LPCSTR,LPCSTR,UINT,WORD,DWORD timeout); PMSGBOXT func; HMODULE hm; UINT t; int ret; hm=::LoadLibrary("user32.dll"); if(!hm){ puts("load DLL error!"); return 1; } func=(PMSGBOXT)::GetProcAddress(hm,"MessageBoxTimeoutA"); if(!func){ puts("get Proc error!"); ::FreeLibrary(hm); return 2; } t=MB_OK|MB_ICONEXCLAMATION|MB_OK; SYSTEMTIME syst; ::GetSystemTime(&syst); c(true); printf("",syst.wHour,syst.wMinute,syst.wSecond,syst.wMilliseconds); c(true); system("start https://www.bilibili.com/video/BV1gJ411W7e8/?spm_id_from=333.337.search-card.all.click"); SetCursorPos(0,0); ret=func(NULL,"Can you hear the silence?","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"Can you see the dark?","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"Can you fix the broken?","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"Can you feel... can you feel my heart?","",t,0,11000); SetCursorPos(0,0); ret=func(NULL,"Can you help the hopeless?","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"Well, I'm begging on my knees,","",t,0,5000); SetCursorPos(0,0); ret=func(NULL,"Can you save my bastard soul?","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"Will you wait for me?","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I'm sorry brothers,","",t,0,1000); SetCursorPos(0,0); ret=func(NULL,"So sorry lover,","",t,0,2000); SetCursorPos(0,0); ret=func(NULL,"Forgive me father,","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"I love you mother.","",t,0,2000); SetCursorPos(0,0); ret=func(NULL,"Can you hear the silence?","",t,0,2000); SetCursorPos(0,0); ret=func(NULL,"Can you see the dark?","",t,0,2000); SetCursorPos(0,0); ret=func(NULL,"Can you fix the broken?","",t,0,2000); SetCursorPos(0,0); ret=func(NULL,"Can you feel my heart?","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"Can you feel my heart?","",t,0,8000); SetCursorPos(0,0); ret=func(NULL,"Can you feel my heart?","",t,0,7000); SetCursorPos(0,0); ret=func(NULL,"Can you feel my heart?","",t,0,16000); SetCursorPos(0,0); ret=func(NULL,"I'm scared to get close and I hate being alone.","",t,0,16000); SetCursorPos(0,0); ret=func(NULL,"I long for that feeling to not feel at all.","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"The higher I get, the lower I'll sink.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I can't drown my demons, they know how to swim.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I'm scared to get close and I hate being alone.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I long for that feeling to not feel at all.","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"The higher I get, the lower I'll sink.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I can't drown my demons, they know how to swim.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I'm scared to get close and I hate being alone.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I long for that feeling to not feel at all.","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"The higher I get, the lower I'll sink.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"I can't drown my demons, they know how to swim.","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"Can you feel my heart?","",t,0,21000); SetCursorPos(0,0); ret=func(NULL,"Can you hear the silence?","",t,0,2000); SetCursorPos(0,0); ret=func(NULL,"Can you see the dark?","",t,0,4000); SetCursorPos(0,0); ret=func(NULL,"Can you fix the broken?","",t,0,3000); SetCursorPos(0,0); ret=func(NULL,"Can you feel... can you feel my heart?","",t,0,4000); SetCursorPos(0,0); ::GetSystemTime(&syst); printf("",syst.wHour,syst.wMinute,syst.wSecond,syst.wMilliseconds); ::FreeLibrary(hm); c(false); return 0; }

快点这儿！

点我 人数：264 本团团长为复仇者_X，有问题可以私信

第一题 思路 > 直接用<cmath>中的abs函数求绝对值即可，或者用 0−n0-n0−n 的减法，取绝对值。 代码 第二题 思路 > 要求4和6的倍数，其实也就是4和6的最小公倍数即12，所以直接判断是否是12的倍数即可。 代码 第三题 思路 > 需要求当前值加上自己的下标即可，这里面不需要用数组存储，直接边输入边输出即可。 代码 第四题 思路 > 遍历当前字符串，如果字符串下标是奇数，则输出 S2[i]S_2[i]S2 [i] ,否则输出 S1[i]S_1[i]S1 [i] ，由于题目中下标是从1开始数的，正常字符串是从0开始数的，所以需要变成为偶数下标时，输出则输出 S2[i]S_2[i]S2 [i] 代码 第五题 思路 > 题目中：“「边缘元素」的定义： 假设 ai,ja_{i,j}ai,j 是边缘元素只要满足 i=1i=1i=1 或者 j=1j=1j=1 或者 i=ni=ni=n 或者 j=nj=nj=n。” > 将上面的条件翻译成代码即可。 代码 第六题 思路 > 考察最大公约数的求法，这里使用辗转相除法。 代码

到底保留几位小数来着？

//体育课下课满头大汗，理物课上课风轻云淡，纤纤素手如电钻，皮肤猛搓发糜烂 //炼化只见顷刻见仙丹，奇技淫巧揉成泥巴腌，触感如超轻粘土，制作手法却简朴 //搓一搓，揉成球，弹遍整个教学楼，搓泥郎，全家唐，同学急眼直骂娘

与上面给的数据差异非常大。