Algorithm
Problem link- https://www.spoj.com/problems/DONOTHIN/
DONOTHIN - Doing Nothing
Original statement in spanish at http://www.dc.uba.ar/events/icpc/download/problems/taip2011-problems.pdf
A young couple uses to make their time as productive as possible. This activity is quite stressful, so they decided to "waste" some time watching their favorite TV series.
The series has N seasons, and each season has a possibly different number of chapters according to its success, actors availability, production time and other external factors. Each chapter has a duration of exactly M minutes.
To keep up with the plot, before watching each new season, they watch, without any rest, all the chapters of all the previous seasons. This has just made them concern about how much time they will be spending with this hobby, which should keep them calm. They need your help so they don't get back to the stressful situation they had.
Input
The input contains several test cases. Each test case is described in two lines. The first line has two integers N and M representing respectively how many seasons the series has and the duration in minutes of each chapter (1 <= N <= 105, 1 <= M <= 106). The next line has N integers C_i representing how many chapters each season has sorted chronologically. (1 <= C_i <= 100 for 1 <= i <= N). The last line of the input contains the number -1 twice and should not be processed as a test case.
Output
For each test case output a single line with an integer representing the number of minutes the couple spends in watching the whole series.
Example
Input:
6 20 24 23 15 22 24 17 1 100 100 10 1000000 99 99 99 99 99 99 99 99 99 99 -1 -1
Output:
9000 10000 5445000000
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <map>
#include <list>
using namespace std;
int main(){
long long n,m,total;
while(true){
total=0;
scanf("%lld%lld",&n,&m);
if(n==-1&&m==-1)break;
while(n--){
//cout<<n<<endl;
long long temp;
scanf("%lld",&temp);
temp=temp*m*(n+1);
total+=temp;
}
printf("%lld\n",total);
}
return 0;
}Input
24 23 15 22 24 17
1 100
100
10 1000000
99 99 99 99 99 99 99 99 99 99
-1 -1
Output
10000
5445000000
Demonstration
SPOJ Solution-Doing Nothing-Solution in C, C++, Java, Python