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Saturday, September 7, 2013
Problem - Pass the time with maximum fun after movie - Solution
Question :
Joey just got free from a shooting and he wants to have a lunch. Director of the movie gave him only "K" units of time for the lunch. There are "N" restaurants in the city. For the ith restuarant, there are two values that Joey knows, "Fi" and "Ti". Ti is the amount of time it would take if Joey eats the lunch in the ith restuarant. If Ti is less than or equal to K, Joey will have Fi amount of fun in that restuarant. If Ti is greater than K, then Joey's fun will be calculated as Fi - (Ti - K).
Joey has the list of all restaurants and Fi and Ti values for all of them. Please tell him the Maximum fun that he can have. He will have his lunch in only one restaurant. The maxiumum can be negative also.
Joey just got free from a shooting and he wants to have a lunch. Director of the movie gave him only "K" units of time for the lunch. There are "N" restaurants in the city. For the ith restuarant, there are two values that Joey knows, "Fi" and "Ti". Ti is the amount of time it would take if Joey eats the lunch in the ith restuarant. If Ti is less than or equal to K, Joey will have Fi amount of fun in that restuarant. If Ti is greater than K, then Joey's fun will be calculated as Fi - (Ti - K).
Joey has the list of all restaurants and Fi and Ti values for all of them. Please tell him the Maximum fun that he can have. He will have his lunch in only one restaurant. The maxiumum can be negative also.
Inputs
First Line will contain two integers N and K. Each of the next N lines will contain two integers Fi and Ti.
First Line will contain two integers N and K. Each of the next N lines will contain two integers Fi and Ti.
Outputs :
Print a single integer that tells the maximum fun that Joey can have.
Print a single integer that tells the maximum fun that Joey can have.
Constraints
1 <= N <= 10^4, 1 <= K <=10^9, 1 <= Fi,Ti <=10^9
1 <= N <= 10^4, 1 <= K <=10^9, 1 <= Fi,Ti <=10^9
Level :Starter,Cakewalk
Solution to the Problem :
#include<stdio.h>
int arr[10001];
void preprocess()
{
arr[1]=10;
int i;
for(i=2;i<=10000;i++)
arr[i]=(arr[i-1]*10)%21;
}
void print(int k)
{
//printf("%d\n",arr[k-2]);
int x=arr[k-2];
printf("1");
int i;
for(i=1;i<=k-4;i++)
printf("0");
if(x==0)
printf("000");
else
{
x=21-x;
if(x<10)
{
printf("0%d0",x);
}
else
{
printf("%d0",x);
}
}
printf("\n");
}
main()
{
preprocess();
int k;
scanf("%d",&k);
if(k==1||k==2)
printf("-1\n");
else if(k==3)
printf("210\n");
else if(k==4)
printf("1050\n");
else
print(k);
}
