using namespace std;
int MyLike(int suiji)
{int donot[10] = {0,1,19,24};
vector<int> notnot ;
//= new vector<int>;
notnot.push_back(0);
notnot.push_back(1);
notnot.push_back(19);
notnot.push_back(24);
vector<int> :: iterator MyItr;
MyItr = notnot.begin();
int L = 0;
for (;MyItr != notnot.end();++MyItr)
{++L;}
/*int K = 0;
for (int j = 0;j < sizeof(donot)/sizeof(int);j++)
{if (donot[j] != NULL)//这边有个问题donot[0] == 0 不走这个判断 因为NULL的宏定义为0{int temp = donot[j];++K;}}*/
int H = notnot.size();
//for (int i=0;i < L;i++)
for (int i=0;i < notnot.size();i++)
{if (suiji != donot[i])
{continue;}
else{suiji = suijishu();
MyLike(suiji);}
}
return suiji;
}
只抠出来了一个函数
可以读读
有想干码农的不
int MyLike(int suiji)
{int donot[10] = {0,1,19,24};
vector<int> notnot ;
//= new vector<int>;
notnot.push_back(0);
notnot.push_back(1);
notnot.push_back(19);
notnot.push_back(24);
vector<int> :: iterator MyItr;
MyItr = notnot.begin();
int L = 0;
for (;MyItr != notnot.end();++MyItr)
{++L;}
/*int K = 0;
for (int j = 0;j < sizeof(donot)/sizeof(int);j++)
{if (donot[j] != NULL)//这边有个问题donot[0] == 0 不走这个判断 因为NULL的宏定义为0{int temp = donot[j];++K;}}*/
int H = notnot.size();
//for (int i=0;i < L;i++)
for (int i=0;i < notnot.size();i++)
{if (suiji != donot[i])
{continue;}
else{suiji = suijishu();
MyLike(suiji);}
}
return suiji;
}
只抠出来了一个函数
可以读读
有想干码农的不