void twoSingleNum(const int *a,int cnt){int tmp=0;for(int i=0;i<cnt;++i)  tmp^=a[i];//比较短的时候放后面int mark=1;while( (tmp&mark)==0)mark<<=1;// mark<<1 has no effect on markint num1=0,num2=0;for(int i=0;i<cnt;++i)(a[i]&mark)==0 ? num1^=a[i] : num2^=a[i];//must have (), &'s priority is very low.cout<<num1<<" "<<num2<<endl;
}

void delChars(char s[],const char *chars){int hash[256];memset(hash,0,sizeof(int)*256);while(*chars!='\0'){hash[int(*chars)]=1;chars++;}char *pFast=s,*pSlow=s;while(*pFast!='\0'){if(hash[int(*pFast)]==0){*pSlow=*pFast;pSlow++;}pFast++;//相当于把不删的复制到前面的slow，删的就跳过
    }*pSlow='\0';
}

void generateUglyNum(int n){int buf[n+10];buf[0]=1;buf[1]=2;buf[2]=3;buf[3]=4;buf[4]=5;int *p2=buf,*p3=buf,*p5=buf;for(int i=5;i<n;++i){while(*p2*2<=buf[i-1])p2++;unsigned int c1=*p2*2;while(*p3*3<=buf[i-1])*p3++;unsigned int c2=*p3*3;while(*p5*5<=buf[i-1])*p5++;unsigned int c3=*p5*5;if(c1<=c2 && c1<=c3)buf[i]=c1;if(c2<=c1 && c2<=c3)buf[i]=c2;if(c3<=c1 && c3<=c2)buf[i]=c3;cout<<i+1<<":"<<buf[i]<<endl;}

void helper(int n){int tmp=0;for(int i=0;i<n;++i){tmp+=9;if(i!=n-1)tmp*=10;}for(int i=1;i<=tmp;++i){cout<<i<<" ";}cout<<endl;
}

第六十六题：利用递归翻转一个栈

template <class T>
void reverse_stack_recursive(stack<T> &s){if(!s.empty()){T tmp=s.top();s.pop();reverse_stack_recursive(s);//从顶到底一个一个拿开putToButtom(s,tmp);//从栈最低的元素一个一个放回，但是每次都返回栈底
    }
}template <class T>//这是一种栈的常用操作，先拿出来，再放回去
void putToButtom(stack<T> &s,T &tmp){if(s.empty()){s.push(tmp);//拿到没有了，就放这个目标元素到栈底return;}T o=s.top();//将栈里的元素一个一个拿开
    s.pop();putToButtom(s,tmp);//直到栈为空，放进目标元素s.push(o);//再一个一个放回去
}

第六十七题：扑克牌顺序

int has_gap(int a[],int l,int allowGaps){for(int i=1;i<5-l;++i){//不同王数for的长度不一，这里用一个参数来传递for的长度就可以实现代码重用，不用写3个for了。if(a[i]-a[i-1]==0)return 0;allowGaps-=a[i]-a[i-1]-1;}if(allowGaps<0)return 0;return 1;
}int is_sequence_good(int a[]){//14 is the kingsort(a,a+5);if(a[3]==14 && a[4]==14)return has_gap(a,2,2);//double kingselse if(a[4]==14)return has_gap(a,1,1);else return has_gap(a,0,0);
}

第六十七题：打印出n粒骰子的点数和的出现概率

void showDicePro(int n){#define DICE_NUM 6 //为以后扩展做准备，注意扩展性可以给面试官好印象int p[n][DICE_NUM*n+1];memset(p,0,sizeof(int)*(DICE_NUM*n+1)*n);for(int j=1;j<=DICE_NUM;++j) p[0][j]=1;//初始化1颗骰子的情况for(int i=1;i<n;++i)for(int j=1;j<=DICE_NUM*n;++j)for(int k=j-DICE_NUM;k<j;++k)if(k>=0) p[i][j]+=p[i-1][k];//n粒的情况DP n-1的for(int i=1;i<=DICE_NUM*n;++i) cout<<p[n-1][i]<<" ";cout<<endl;
}

第六十八题：把数组组成最小的数

bool fronter(int a,int b){//思路1vector<int> va,vb,tmp;while(a!=0){tmp.push_back(a%10);a/=10;}while(!tmp.empty()){va.push_back(tmp.back());tmp.pop_back();}while(b!=0){tmp.push_back(b%10);b/=10;}while(!tmp.empty()){vb.push_back(tmp.back());tmp.pop_back();}//比较长度int lena=va.size();int lenb=vb.size();int i=0,j=0;while(i<lena || j<lenb){if(va[i]<vb[j] && i<lena && j<lenb)return true;if(va[i]>vb[j] && i<lena && j<lenb)return false;if(i==lena && j==lenb)return false;if(i==lena && j!=lenb){//前面的放前面就是trueif(vb[j]>vb[j-lena])return true;//短在前else if(vb[j]<vb[j-lena]) return false;}//多出来的位与第一位比if(i!=lena && j==lenb){if(va[i]>va[i-lenb])return false;//短在前else if(va[i]<va[i-lenb])return true;}if(i<lena)++i;if(j<lenb)++j;}return true;
}bool helper(int a,int b){//思路2char sa[50],sb[50];sprintf(sa,"%d",a);sprintf(sb,"%d",b);string stra(sa),strb(sb);return stra+strb<strb+stra;
}void final(int *num,int n){vector<int> a;for(int i=0;i<n;++i)a.push_back(num[i]);sort(a.begin(),a.end(),helper);for(unsigned int i=0;i<a.size();++i){cout<<a[i];}cout<<endl;
}