Saturday 29 August 2020

Weather News Website

 

Prerequisite-

1.HTML

2.CSS

3.Javascript

Characteristics of Website-

Current Temperature,Max Temperature,Location,Description,Wind Speed.

How it Work-

We fetch data from API 

Step 1-

Create account on https://openweathermap.org/ for Api.

step 2-

Show json data .

Javascript code-

<script type="text/javascript">
window.addEventListener("load",()=>{
let long;
let lat;
if(navigator.geolocation)
{
navigator.geolocation.getCurrentPosition(position=>{
long=position.coords.longitude;
late=position.coords.latitude;
const Api="https://api.openweathermap.org/data/2.5/weather?lat="+late+"&lon="+long+"&appid=API KEY";
//console.log(Api);
fetch(Api)
.then(response=>{
return response.json();
})
.then(data=>{
var dt=data;
console.log(dt);
var json_object=JSON.parse(JSON.stringify(dt));
var x=parseInt(json_object.main.temp);
var y=(parseInt(x)-273.15);
document.getElementById("tmp").innerHTML="Temparature "+y+"`c";
var x=parseInt(json_object.main.temp_max);
var y=(parseInt(x)-273.15);
document.getElementById("mxTmp").innerHTML="Max Temp "+y+"`c";
 
document.getElementById("des").innerHTML="Description-"+json_object.weather[0].description;
document.getElementById("wdSpeed").innerHTML="wind Speed -"+data.wind.speed;
document.getElementById("loc").innerHTML="Current Weather in "+data.name+"/"+data.sys.country;
 
});
});
}
});
</script>>

CSS code-

<style>
body {
margin: 100px 100px 75px 100px;
background-color: coral;
background: linear-gradient(rgb(100,240,100),rgb(48,62,143));
}
h1{
font-size: 15;
font-family: sans-serif;
font-style: oblique;
 
}
p{
border-style: double;
}
</style>

Complete code-

</!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<style>
body {
margin: 100px 100px 75px 100px;
background-color: coral;
background: linear-gradient(rgb(100,240,100),rgb(48,62,143));
}
h1{
font-size: 15;
font-family: sans-serif;
font-style: oblique;
 
}
p{
border-style: double;
}
</style>
<script type="text/javascript">
window.addEventListener("load",()=>{
let long;
let lat;
if(navigator.geolocation)
{
navigator.geolocation.getCurrentPosition(position=>{
long=position.coords.longitude;
late=position.coords.latitude;
const Api="https://api.openweathermap.org/data/2.5/weather?lat="+late+"&lon="+long+"&appid=abb0c011119d36f2fe4d4244b91955d2";
//console.log(Api);
fetch(Api)
.then(response=>{
return response.json();
})
.then(data=>{
var dt=data;
console.log(dt);
var json_object=JSON.parse(JSON.stringify(dt));
var x=parseInt(json_object.main.temp);
var y=(parseInt(x)-273.15);
document.getElementById("tmp").innerHTML="Temparature "+y+"`c";
var x=parseInt(json_object.main.temp_max);
var y=(parseInt(x)-273.15);
document.getElementById("mxTmp").innerHTML="Max Temp "+y+"`c";
 
document.getElementById("des").innerHTML="Description-"+json_object.weather[0].description;
document.getElementById("wdSpeed").innerHTML="wind Speed -"+data.wind.speed;
document.getElementById("loc").innerHTML="Current Weather in "+data.name+"/"+data.sys.country;
 
});
});
}
});
</script>>
<svg width="1em" height="1em" viewBox="0 0 16 16" class="bi bi-sun" fill="currentColor" xmlns="http://www.w3.org/2000/svg">
<path d="M3.5 8a4.5 4.5 0 1 1 9 0 4.5 4.5 0 0 1-9 0z"/>
<path fill-rule="evenodd" d="M8.202.28a.25.25 0 0 0-.404 0l-.91 1.255a.25.25 0 0 1-.334.067L5.232.79a.25.25 0 0 0-.374.155l-.36 1.508a.25.25 0 0 1-.282.19l-1.532-.245a.25.25 0 0 0-.286.286l.244 1.532a.25.25 0 0 1-.189.282l-1.509.36a.25.25 0 0 0-.154.374l.812 1.322a.25.25 0 0 1-.067.333l-1.256.91a.25.25 0 0 0 0 .405l1.256.91a.25.25 0 0 1 .067.334L.79 10.768a.25.25 0 0 0 .154.374l1.51.36a.25.25 0 0 1 .188.282l-.244 1.532a.25.25 0 0 0 .286.286l1.532-.244a.25.25 0 0 1 .282.189l.36 1.508a.25.25 0 0 0 .374.155l1.322-.812a.25.25 0 0 1 .333.067l.91 1.256a.25.25 0 0 0 .405 0l.91-1.256a.25.25 0 0 1 .334-.067l1.322.812a.25.25 0 0 0 .374-.155l.36-1.508a.25.25 0 0 1 .282-.19l1.532.245a.25.25 0 0 0 .286-.286l-.244-1.532a.25.25 0 0 1 .189-.282l1.508-.36a.25.25 0 0 0 .155-.374l-.812-1.322a.25.25 0 0 1 .067-.333l1.256-.91a.25.25 0 0 0 0-.405l-1.256-.91a.25.25 0 0 1-.067-.334l.812-1.322a.25.25 0 0 0-.155-.374l-1.508-.36a.25.25 0 0 1-.19-.282l.245-1.532a.25.25 0 0 0-.286-.286l-1.532.244a.25.25 0 0 1-.282-.189l-.36-1.508a.25.25 0 0 0-.374-.155l-1.322.812a.25.25 0 0 1-.333-.067L8.203.28zM8 2.5a5.5 5.5 0 1 0 0 11 5.5 5.5 0 0 0 0-11z"/>
</svg>
<div>
 
<h2 id="loc">Location</h2>
</div>
<div class="double">
<h1 id="tmp">Temparature</h1>
<h1 id="mxTmp">Max Temp</h1>
 
<h1 id="des">Description</h1>
<h1 id="wdSpeed">wind Speed</h1>
</div>
<div id="anim">
</div>
 
</body>
</html>>



at 5 comments:

Saturday 22 August 2020

CSES Dynamic Programming Problem Solution+Explanation

Problem 1-Dice Combinations

Editorial-

Now start from base 
1->1
2->1+1,2+0
3->1+1+1,2+1,1+2,3+0
now for-4
we substract from 1 to min(6,4)
( we substract 1 then remaining 3 now we add no.of possible way to constuct 3)+(we substract 2 remaining(4-2=2,no of possible way to constuct 2)+(we substract 3,no. of possible way to construct 1)+(we substract 4 ,now of possible way to construct 0).
i think you got the point
base condition-
dp[0]=1;
how dp[0] ,for this there is empty subset exist

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
#define mod 1000000007
ll dp[1000010];
 
int main()
{
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin>>n;
dp[0]=1;
for(int i=1;i<=n;i++)
{
ll ans=0;
for(int j=1;j<=min(6,i);j++)
{
ans+=dp[i-j];
ans%=mod;
}
dp[i]=ans;
 
}
cout<<dp[n];
 
}

Problem 2-Minimizing Coins

Editorial-

Parent Problem-Unbounded Knapsack

Let understand through example-
3 11
1 5 7
we have to construct 11 from minimum no of coins from given coin {1,5,7}
minimum coins for 1->1
minimum coins for 2->1+1
minimum coins for 3->min(1 coin of 1+(3-1))->1 coin of 1 +minimum  coins for construct 2
we have already calculated minimum coins for 2
dp[3]=1+dp[2-1]
minimum coin for 4->1 coin of 1+minimum coins for construct 3
minimum coins for 5->{1 coins of 1+minimum coins for construct 4} or {1 coin of 5 +minimum coins for construct 0}
and so on. 
dp[i]=minimum coin required to constuct i.
base consition-
dp[0]=0

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
ll dp[1000010];
 
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n,x;
cin>>n>>x;
ll a[n];
for(int i=0;i<n;i++)
cin>>a[i];
dp[0]=0;
for(int i=1;i<=x;i++)
{ dp[i]=INT_MAX;
for(int k=0;k<n;k++)
{ if(a[k]<=i)
{
dp[i]=min(dp[i],1+dp[i-a[k]]);
}
}
}
if(dp[x]==INT_MAX)
cout<<"-1";
else
cout<<dp[x];
}

Problem 3-Coin Combinations I

Editorial-

This is same as Previous in this problem we have to add instead of find minimum
Eg.
3 9
2 3 5
0->1
1->0
2->1 ways{1}
3->{2,ways{1}}+{3,ways{0}}// we already calculated for 1 and 0
4->{2,ways{4-2=2}}+{3,ways{1}} 
5->{2,ways{5-2=3}}+{3,ways{5-3=2}}+{5.,ways{0}}
and so on....
base condition dp[0]=1;

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
#define mod 1000000007
ll dp[1000010];
 
int main()
{
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n,x;
cin>>n>>x;
ll a[n];
for(int i=0;i<n;i++)
cin>>a[i];
dp[0]=1;
for(int i=1;i<=x;i++)
{
for(int j=0;j<n;j++)
{
if(a[j]<=i)
{
dp[i]+=dp[i-a[j]];
dp[i]%=mod;
}
}
// cout<<dp[i]<<" ";
}
cout<<dp[x];
}

Problem 4-Coin Combinations II

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
#define mod 1000000007
 
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n,x;
cin>>n>>x;
int a[n];
vector<vector<int>> dp(n+1,vector<int>(x+1,0));
dp[0][0]=1;
 
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=1;i<=n;i++)
{
for(int j=0;j<=x;j++)
{ dp[i][j]=dp[i-1][j];
if(a[i-1]<=j)
{
(dp[i][j]+=dp[i][j-a[i-1]])%mod;
dp[i][j]%=mod;
}
}
}
cout<<dp[n][x];
}

Problem 5-Removing Digits

Editorial-

This problem is based on 0-1 knapsack
for example 
                   we remove each digit and take minimum of each case
But there is a deadlock if we substract 0 so avoid 0.

Technique-recursion+memoization

Solution-
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
int dp[1000010];
int solve(int n)
{ if(n<0)
return 1e8;
if(n==0)
return 0;
if(dp[n]!=-1)
return dp[n];
int ans=1e8,p=n;
while(p!=0)
{
ll x=p%10;
p/=10;
if(x!=0)
ans=min(ans,1+solve(n-x));
}
return dp[n]=ans;
}
int main()
{
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin>>n;
memset(dp,-1,sizeof(dp));
solve(n);
cout<<dp[n];
}

Problem 6-Grid Paths

Editorial- 

base case -
row 0-if there is a block then from that block to n=0
column 0-if there is block then from that block 0 to n=0

otherwise if
there is no "#" in left and up then dp[i][j]=dp[i-1][j]+dp[i][j-1]



Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
#define mod 1000000007
ll dp[1001][1001];
 
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin>>n;
string s[n+1];
for(int i=0;i<n;i++)
cin>>s[i];
if(s[0][0]=='*')
dp[0][0]=0;
else
dp[0][0]=1;
for(int i=1;i<n;i++)
{
if(s[i][0]=='*')
dp[i][0]=0;
else
dp[i][0]=dp[i-1][0];
if(s[0][i]=='*')
dp[0][i]=0;
else
dp[0][i]=dp[0][i-1];
}
for(int i=1;i<n;i++)
{
for(int j=1;j<n;j++)
{
if(s[i][j]=='*')
{
dp[i][j]=0;
continue;
}
if(s[i-1][j]==s[i][j-1]&&s[i-1][j]!='*')
dp[i][j]=(dp[i-1][j]+dp[i][j-1])%mod;
else
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
dp[i][j]%=mod;
 
}
}
cout<<dp[n-1][n-1];
}

Problem 7-Book Shop

Editorial-

Based on 0-1 knapsack.

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
 
int main()
{
int n,x;
cin>>n>>x;
vector<int>p(n),v(n);
for(int i=0;i<n;i++)
cin>>p[i];
for(int i=0;i<n;i++)
cin>>v[i];
vector<vector<int>> dp(n+1,vector<int>(x+1,0));
for(int i=1;i<=n;i++)
{
for(int j=0;j<=x;j++)
{
dp[i][j]=dp[i-1][j];
if(j-p[i-1]>=0)
dp[i][j]=max(dp[i][j],v[i-1]+dp[i-1][j-p[i-1]]);
}
}
cout<<dp[n][x];
}

Problem 8-Array Description

Editorial-

For every xi=0 we fill it from 1 to m and check is it valid or not in each step we take prev and pointer i
if prev-xi>1 we return 0
and if i reach n then this will we valid combination so return 1
and sum all these combition will be answer.
Technique- recursion+memoization

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
#define mod 1000000007
ll n,m,a[100001];
ll dp[100001][101];
 
ll solve(ll curr,ll i)
{
if(i==n)
return 1;
if(dp[i][curr]!=-1)
return dp[i][curr];
if(a[i]!=0)
{
if(abs(curr-a[i])>1)
return 0;
else
return solve(a[i],i+1);
}
else
{
ll ans=0;
for(int j=1;j<=m;j++)
{
if(abs(j-curr)<=1)
ans+=solve(j,i+1);
ans%=mod;
}
 
return dp[i][curr]=ans;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin>>n>>m;
memset(dp,-1,sizeof(dp));
for(int i=0;i<n;i++)
cin>>a[i];
if(a[0]!=0)
cout<<solve(a[0],1);
else
{ ll ans=0;
for(int i=1;i<=m;i++)
{
ans+=solve(i,1);
ans%=mod;
}
cout<<ans;
}
}

Problem 9-Edit Distance

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
 
 
int main()
{
 
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s1,s2;
cin>>s1>>s2;
int n=s1.size(),m=s2.size();
vector<vector<int>> dp(n+1, vector<int>(m+1,1e9));
dp[0][0]=0;
for(int i=0;i<=n;i++)
{
for(int j=0;j<=m;j++)
{ if (i) {
dp[i][j] = min(dp[i][j], dp[i-1][j]+1);
}
if (j) {
dp[i][j] = min(dp[i][j], dp[i][j-1]+1);
}
if (i && j) {
dp[i][j] = min(dp[i][j], dp[i-1][j-1]+(s1[i-1] != s2[j-1]));
}
}
}
cout<<dp[n][m];
}

Problem 10-Rectangle Cutting

Editorial-

In each cut we divide into 2 
for example 
3*4 can be-{3*1+3*3} or {3*2+3*2} or {3*3+3*1}
or
3*4={1*4+2*4} or{2*4+1*4}
squere if i==j
then minimum cut =0;

Solution-

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define str string
#define pb push_back
#define vc vector
#define ci cin
#define co cout
ll dp[1000][1000];
 
int main()
{
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{ if(i==j)
dp[i][j]=0;
else
{ dp[i][j]=INT_MAX;
for(int k=1;k<j;k++)
dp[i][j]=min(dp[i][j],(1LL+dp[i][k]+dp[i][j-k]));
for(int k=1;k<i;k++)
dp[i][j]=min(dp[i][j],(1LL+dp[i-k][j]+dp[k][j]));
}
}
}
cout<<dp[n][m];
}

If there is a mistake then tell me i will correct .
Happy Coding.
at No comments:
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