From abe1c2ffb563700b9524f4f2f6ef173ed1a37dd9 Mon Sep 17 00:00:00 2001 From: Ankit Srivastava <42239246+ankitsri98@users.noreply.github.com> Date: Tue, 22 Oct 2019 21:42:33 +0530 Subject: [PATCH 1/3] Create boredom --- src/boredom | 67 +++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 67 insertions(+) create mode 100644 src/boredom diff --git a/src/boredom b/src/boredom new file mode 100644 index 0000000..eb36319 --- /dev/null +++ b/src/boredom @@ -0,0 +1,67 @@ +A. Boredom +time limit per test1 second +memory limit per test256 megabytes +inputstandard input +outputstandard output +Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it. + +Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player. +Alex is a perfectionist, so he decided to get as many points as possible. Help him. +Input +The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence. + +The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105). + +Output +Print a single integer — the maximum number of points that Alex can earn. + +Examples +inputCopy +2 +1 2 +outputCopy +2 +inputCopy +3 +1 2 3 +outputCopy +4 +inputCopy +9 +1 2 1 3 2 2 2 2 3 +outputCopy +10 +Note +Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. + +SOLUTION :- + +#include +using namespace std; + +int main() +{ + int n,x; + cin>>n; + long long int a[100005]={0}; + long long int dp[100005]={0}; + + for(int i=0;i>x; + a[x]++;// cREATING A FREQUENCY ARRAY USING ITS INDEX AS NUMBERS + } + dp[0]=0; //if INPUT ARRAY ONLY HAS 0 + dp[1]=a[1];//if input ARRAY ONLY HAS 0 AND 1'S + for(int i=2;i<=100000;i++) + { + dp[i]=max(dp[i-1],dp[i-2]+(i*a[i]));//IF NOT TAKING ELMENT THEN TAKE THE PREVIOUS ONE + //ELSE TAKE THE PREVIOUS TO PREVIOUS AND ADD THE FREQ OF PRESENT ELEMENT + } + /* for(int i=0;i<=1000;i++) + { + cout< Date: Wed, 30 Oct 2019 20:42:50 +0530 Subject: [PATCH 2/3] Create spoj dp problem --- src/spoj dp problem | 1 + 1 file changed, 1 insertion(+) create mode 100644 src/spoj dp problem diff --git a/src/spoj dp problem b/src/spoj dp problem new file mode 100644 index 0000000..7085d28 --- /dev/null +++ b/src/spoj dp problem @@ -0,0 +1 @@ +#famous easy dp problem From 5be7d9c03e914b4eb604843862c702b47365c1fc Mon Sep 17 00:00:00 2001 From: Ankit Srivastava <42239246+ankitsri98@users.noreply.github.com> Date: Wed, 30 Oct 2019 20:43:40 +0530 Subject: [PATCH 3/3] Update spoj dp problem --- src/spoj dp problem | 67 +++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 67 insertions(+) diff --git a/src/spoj dp problem b/src/spoj dp problem index 7085d28..88d2ce3 100644 --- a/src/spoj dp problem +++ b/src/spoj dp problem @@ -1 +1,68 @@ #famous easy dp problem +A. Boredom +time limit per test1 second +memory limit per test256 megabytes +inputstandard input +outputstandard output +Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it. + +Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player. +Alex is a perfectionist, so he decided to get as many points as possible. Help him. +Input +The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence. + +The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105). + +Output +Print a single integer — the maximum number of points that Alex can earn. + +Examples +inputCopy +2 +1 2 +outputCopy +2 +inputCopy +3 +1 2 3 +outputCopy +4 +inputCopy +9 +1 2 1 3 2 2 2 2 3 +outputCopy +10 +Note +Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. + +SOLUTION :- + +#include +using namespace std; + +int main() +{ + int n,x; + cin>>n; + long long int a[100005]={0}; + long long int dp[100005]={0}; + + for(int i=0;i>x; + a[x]++;// cREATING A FREQUENCY ARRAY USING ITS INDEX AS NUMBERS + } + dp[0]=0; //if INPUT ARRAY ONLY HAS 0 + dp[1]=a[1];//if input ARRAY ONLY HAS 0 AND 1'S + for(int i=2;i<=100000;i++) + { + dp[i]=max(dp[i-1],dp[i-2]+(i*a[i]));//IF NOT TAKING ELMENT THEN TAKE THE PREVIOUS ONE + //ELSE TAKE THE PREVIOUS TO PREVIOUS AND ADD THE FREQ OF PRESENT ELEMENT + } + /* for(int i=0;i<=1000;i++) + { + cout<