;//求x!中k因数的个数。int Grial(int x,int k){int Ret = 0;while (x){Ret += x / k;x /= k;}return Ret;}int main(){cout << Grial(10, 2) << endl;return 0;};int Grial(int x, int k){int count = 0;int n = x;while (n){n &= (n – 1);count++;}return x – count;}int main(){cout << Grial(3, 2) << endl;return 0;};int Grial(int a[], int n){int count=0;int key;for (int i = 0; i < n; i++){if (count == 0){key = a[i];count = 1;}else{if (key == a[i]){count++;}else{count–;}}}return key;}int main(){int a[] = {1,2,3,4,5,6,3,3,3,3,3};cout<<Grial(a, sizeof(a) / sizeof(int))<<endl;return 0;};void Grial(int a[], int n){if (n <= 3)return;int count1=0, key1=0;int count2=0, key2=0;int count3=0, key3=0;for (int i = 0; i < n; i++){if (!count1 && key2 != a[i] && key3 != a[i]){count1++;key1 = a[i];}else if (key1 == a[i]){count1++;}else if (key2!=a[i] && key3!=a[i]){count1–;}if (!count2 &&key3 != a[i] && key1!=a[i]){count2++;key2 = a[i];}else if (key2 == a[i]){count2++;}else if (key1!=a[i] && key3!=a[i]){count2–;}if (!count3 && key1!=a[i] && key2!=a[i]){count3++;key3 = a[i];}else if (key3 == a[i]){count3++;}else if (key1!=a[i] && key2!=a[i]){count3–;}}cout << key1 << endl;cout << key2 << endl;cout << key3 << endl;}int main(){int a[] = {1,5,5,5,5,2,3,1,2,2,1,1,1,2};Grial(a, sizeof(a) / sizeof(int));return 0;}
，对人性的弱点有清醒的认识，
