2026信息素养初赛-数字守艺人-小学4～6年级组C++模拟题3

答案：B在模拟皮影戏角色动作的C++程序中，要让角色完成"抬手-迈步-放下手"的循环动作，每次循环间隔1raise_hand();step();lower_hand();sleep(1);}raise_hand();step();lower_hand();sleep(1);}raise_hand();step();lower_hand();sleep(1);raise_hand();step();lower_hand();sleep(1);}解析：选项A中while(true)是无限循环条件，会持续执行循环体内的动作；选项B的for循环仅执行1A.std::mapB.std::vectorC.std::set答案：B解析：二十四节气是有序的固定列表，std::vector是动态数组，适合存储有序且需要按顺序访问的元素；std::map用于键值对映射，std::set用于无重复的无序集合，std::queue用于队列操作，均不符合有}}}}50，会越界访问数组；选项C的while循环语法错误，不在C++程序中模拟陶艺拉坯的转速调节，转速范围是50-200转/分钟，用户输入的转速如果超出范围则private:public:};B.public:private:};C.private:private:};public:public:};设为私有，不符合封装要求；选项C将获取方法设为私有，外部doublecowRate=(doubleprintf("牛皮占比：%.2f%%\n",cowRate);printf("牛皮占比：%.2f%%\n",cowRate);doublecowRate=(double)(cowhide/100)*1printf("牛皮占比：%.2f%%\n",cowRate);doublecowRate=(doubprintf("牛皮占比：%.2f%%\n",cowRate);1/100。boolisCentralSymmetric(voidisCentralSymmetric(boolpatterboolisCentralSymmetric(boolpattern[5]);选项C的返回值为void，不符合题目需要返回判在C++程序中，用std::vector存储非遗传承人的年龄数doubleavg=(double)sum/ages.size();doubleavg=sum/ages.size();vector<int>::iteratorit;for(it=ages.begin();it!=ages.end();it++)sum+=*it;doubleavg=sum/ages.size();doubleavg=sum/ages.size();数字守艺人场景中，用C++编写程序模拟刺绣的针法顺序，针法序列为"平针-打籽针-回针-平针"，需要stringstitches[]={"平针","打籽针","回针"}}stringstitches[]={"平针","打籽针","回针"}}}stringstitches[]={"平针","打籽针","回针"}sort(works,works+100,[](constWork&a,constWork&b){returna.getTime()>b.getTime();sort(works,works+100,[](constWork&a,constWork&b){returna.getTime()<b.getTime();sort(works.begin(),works.end(),[](constWork&a,constWork&b){returna.getTime()>b.getTime();sort(works.begin(),works.end(),[](constWork&a,constWork&b){returna.getTime()<b.getTime();A.std::vectorB.std::mapC.std::listD.std::stack答案：B解析：节气名称和农事活动是键值对的映射关系，ststd::vector、std::list是线性容器，查找效率低；std::stack是栈结构，不适合键值对查找。vector<pair<int,int>>path;path.push_back(make_pair(0,0));path.push_back(make_pair(10,0));path.push_back(make_pair(10,10));path.push_back(make_pair(0,10));path.push_back(make_pair(0,0));}vector<pair<int,int>>path;path.push_back(make_pair(0,0));path.push_back(make_pair(10,0));path.push_back(make_pair(10,10));path.push_back(make_pair(0,10));pat