The C + + solution array has three digits in the Super 1/4.

#include <iostream>using namespace STD;//x! the number of k factors in the intGrial (intXintK) {intRet =0; while(x) {Ret + = x/k;    x/= k; }returnRet;}intMain () {cout<< Grial (Ten,2) << Endl;return 0;}//If the number of factors in K in a n! is required, then the following rules must be met. //IE x=n/k+n/k^2+n/k^3 ... (until N/k^x is less than 0);#include <iostream>using namespace STD;intGrial (intXintK) {intCount =0;intn = x; while(n) {n &= (n-1);    count++; }returnX-count;}intMain () {cout<< Grial (3,2) << Endl;return 0;}//Find out that the number of occurrences in the array exceeds half the array. #include <iostream>using namespace STD;intGrial (intA[],intN) {intCount=0;intKey for(inti =0; I < n; i++) {if(Count = =0) {key = A[i]; Count =1; }Else{if(key = = A[i])            {count++; }Else{count--; }        }    }returnKey;}intMain () {intA[] = {1,2,3,4,5,6,3,3,3,3,3};cout<<grial (A,sizeof(a)/sizeof(int)) <<endl;return 0;}#include <iostream>//The extension of the previous topic, there are 3 number occurrences more than 1/4. using namespace STD;voidGrial (intA[],intN) {if(N <=3)return;intcount1=0, key1=0;intCount2=0, key2=0;intcount3=0, key3=0; for(inti =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;}intMain () {intA[] = {1,5,5,5,5,2,3,1,2,2,1,1,1,2}; Grial (A,sizeof(a)/sizeof(int));return 0;}

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The C + + solution array has three digits in the Super 1/4.