SoJ 4390 Elevator Problem

Background: The weekly question was not read at the time. Even if you read it, you can only think of violence, not AC.

Learning: 1. In the case of violent search timeouts, a good algorithm must be found, which is to look at the optimal solution with a similar trend of change rather than figure out the corresponding value for each floor and find the maximum value. Train of thought: Assume current floor below has N1 person, current floor has N2 person, current floor above has N3 person. Every step up, there's n1+n2 people going to take the first floor.

People want to walk the first floor, if starting from the No. 0 floor to consider, then N1+N2 is 0,n3 for the total number of people, and then go upstairs, N1+n2 monocytogenes, N3 single minus, obviously N1+n2>=n3 stop upstairs is to take maximum value.

Code:

#include <stdio.h> #include <stdlib.h> #include <math.h>short str[1000000];int main () {    int t;    scanf ("%d", &t);    while (t--) {    int n,floor,n1=0,n2=0,n3,sum=0;    Long long people=0;    scanf ("%d", &n);    for (int i=0;i<n;i++) {        scanf ("%hd", &str[i]);        Sum+=str[i];    }    for (int i=0;i<n;i++) {        if (i>0)        n1+=str[i-1];        N2=str[i];        N3=SUM-N1-N2;        if (n1+n2-n3>=0) {        floor=i+1;        for (int k=0;k<n;k++) people+=str[k]* (ABS (I-K));        break;        }        }        printf ("%d%lld\n", floor,people);    } return 0;}

SoJ 4390 Elevator Problem