Maximum continuous sequence and, product, and disjoint array and the difference between the largest

Problem One: The maximum number of sub-arrays and:

Use F[i] to denote the largest continuous subsequence and at the end of A[i]. I find the maximum value between 0~n-1 and the last comparison of all F[i]. For such a dynamic plan, you can simplify the update by scrolling through a variable F.

intMax_sum (intA[],intN) {    intf=a[0]; intmax_value=a[0];  for(intI=1; i<n;i++)    {        if(f<0) F=A[i]; ElseF=f+A[i]; Max_value=Max (max_value,f); }    returnMax_value;}

Question two: The maximum value of the disjoint subarray and its difference.

We need to divide the array into two disjoint parts, one for each possible two-part sub-array and the maximum value, and the minimum value. For example, the array a[0~n] is divided into a[0~i] and a[i+1~n], we use the f1[i] to represent the first part of the a[0~i] sub-array of the largest continuous sub-sequence and. F2[i] represents the and of the smallest contiguous subsequence of the first part of the a[0~i] sub-array. G1[I+1] represents the largest contiguous subsequence of the second part of the a[i+1~n] sub-array and. G2[i+1] represents the and of the smallest contiguous subsequence of the second part of the a[i+1~n] sub-array. Finally, for each I, we find the difference between the largest two parts, Max (f1[i]-g2[i+1], g1[i+1]-f2[i]), where f1[i],f2[i],g1[i],g2[i] is the result obtained on the basis of problem one;

intMax_diff_sum (intA[],Const intN) {Vector<int> f1 (n);//F1[i] Represents the maximum value of all successive sub-sequences contained in the A[0~i] sub-arrayvector<int>F2 (n); Vector<int>G1 (n); Vector<int>G2 (n); f1[0]=a[0]; f2[0]=a[0]; intmax_left_until=a[0];//represents the maximum number of consecutive sub-sequences ending with a[i]    intmin_left_until=a[0];//represents the minimum value of a continuous subsequence ending with a[i]     for(intI=1; i<n;i++)    {        if(max_left_until>0) Max_left_until+=A[i]; ElseMax_left_until=A[i]; F1[i]=max (f1[i-1],max_left_until); if(min_left_until<0) Min_left_until+=A[i]; ElseMin_left_until=A[i]; F2[i]=min (f2[i-1],min_left_until); } g1[n-1]=a[n-1]; G2[n-1]=a[n-1]; intmax_right_until=a[n-1]; intmin_right_until=a[n-1];  for(inti=n-2; i>=0; i--)    {        if(max_right_until>0) Max_right_until+=A[i]; ElseMax_right_until=A[i]; G1[i]=max (g1[i+1],max_right_until); if(min_right_until<0) Min_right_until+=A[i]; ElseMin_right_until=A[i]; G2[i]=min (g2[i+1],min_right_until); }    intresult=int_min;  for(intI=0; i<n-1; i++) {result=max (Result,max (f1[i]-g2[i+1],g1[i+1]-f2[i])); }    returnresult;}

Question three: maximal continuous product substring:

For a substring of the maximum continuous product of an array, because the product may have a positive negative, if the preceding product is negative, just a[i] is negative, it will be negative. So we need max_until to save the maximum value of the product of the substring ending with a[i], and the minimum value of the product of the substring with Min_until to the end of the a[i]. Then the maximum value in all the Max_until is the final result.

DoubleMax_multi (DoubleA[],intN) {    Doublemax_until=a[0]; Doublemin_until=a[0]; Doubleresult=a[0];  for(intI=1; i<n;++i) {Doubletemp1=max_until*A[i]; Doubletemp2=min_until*A[i]; Max_until=Max (Max (TEMP1,TEMP2), a[i]); Min_until=min (min (TEMP1,TEMP2), a[i]); Result=Max (result,max_until); }    returnresult;}

Maximum continuous sequence and, product, and disjoint array and the difference between the largest