To find the largest contiguous and

If we want to multiply the product and consider the negative number multiplication and the case of 0 ...!

Continuous product sub-series Max

mi[1] = mx[1] = a[1];          for (int2; I <= n;i++) {            = min (min (a[i],a[i]*mi[i-1]), a[i]*mx[i-1 ]);             = Max (max (a[i],a[i] * mi[i-1]), A[i] * mx[i-1]);        }

#include <iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<cstdlib>#include<stack>#include<cctype>#include<string>#include<malloc.h>#include<queue>#include<map>using namespacestd;Const intINF =0xFFFFFF;Const DoubleESP = 10e-8;Const DoublePi =4* ATAN (1.0);Const intMAXN = -+Ten;Const Long LongMoD =2147483647;Const intDr[] = {1,0,-1,0,-1,1,-1,1};Const intDc[] = {0,1,0,-1,1,-1,-1,1};typedefLong LongLL; ll GAC (ll A,ll b) {returnB?GAC (b,a%b): A;}inta[10000];intMaxsum (intXintY) {///the algorithm is complex NLN (n), f (n) = 2 * F (N/2) +n,f (1) = 1;    if(Y-x = =1)        returnA[x]; intm = x + (y-x)/2; intMM =Max (Maxsum (x,m), Maxsum (m,y)); intL = a[m-1]; intR = a[y-1]; intv =0;  for(inti = m1; I >-1; i--) {v+=A[i]; L=Max (v,l); } v=0;  for(inti = M;i < y;i++) {v+=A[i]; R=Max (R,V); }    returnMax (mm,l+R);}intMain () {#ifndef Online_judge freopen ("Inpt.txt","R", stdin);#endif    intN;  while(~SCANF ("%d",&N)) {        ints[10000]; memset (s),0,sizeof(s)); intMX =-0xfffff; inttt =-1;  for(inti =1; I <= n;i++) {scanf ("%d",&A[i]); S[i]+ = (s[i-1] +A[i]); if(S[i] >=MX) {MX=S[i]; TT=i; }        }        intMi = a[1]; if(TT! =-1){             for(inti =2; i < tt;i++) {///look for the minimum value in front of the largest and most foundMi =min (mi,[i]); }} cout<<"D1:"<< Mx-mi <<Endl; intxx = Maxsum (1, n+1); cout<<"D2:"<< xx <<Endl; }    return 0;}

Continuous sub-sequence Max and

To find the largest contiguous and