# Daily Coding Challenge #65

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
Angle Between Hands of a Clock

Given two numbers, hour and minutes. Return the smaller angle (in degrees) formed between the hour and the minute hand.

Example 1:

Input: hour = 12, minutes = 30
Output: 165
Example 2:

Input: hour = 3, minutes = 30
Output: 75
Example 3:

Input: hour = 3, minutes = 15
Output: 7.5
Example 4:

Input: hour = 4, minutes = 50
Output: 155
Example 5:

Input: hour = 12, minutes = 0
Output: 0

Constraints:

1 <= hour <= 12
0 <= minutes <= 59
Answers within 10^-5 of the actual value will be accepted as correct.
*/

class Solution {
public:
double angleClock(int hour, int minutes) {
// each minute is 6 degree (360/60)
// each hour is 30 degree (360/12)
// echo hour has 60 mins so each min is 0.5 degree (30/60)
double minute=minutes*6, hr=hour*30+(double)minutes/2, angle=abs(hr-minute);
// if it is obtuse, make it acute
return min(angle, 360-angle);
}
};


/*
ITMO Academy: pilot course - Segment Tree - A. Segment Tree for the Sum
https://codeforces.com/edu/course/2/lesson/4/1/practice/contest/273169/problem/A
*/

#include <bits/stdc++.h>
using namespace std;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

struct segtree {
int size;
vector<long long> sums;

void init(int n){
size = 1;
while(size < n) size *= 2;
sums.assign(2*size,0LL);
}

void set(int i, int v, int x, int lx, int rx){
if(rx-lx==1){
sums[x]=v;
return;
}

int m = (lx+rx)/2;
if(i<m) set(i,v,2*x+1,lx,m);
else set(i,v,2*x+2,m,rx);
sums[x] = sums[2*x+1]+sums[2*x+2];
}

void set(int i, int v){
set(i, v, 0, 0, size);
}

long long sum(int l, int r, int x, int lx, int rx){
// no intersection
if(lx>=r||l>=rx) return 0;
// inside
if(lx>=l&&rx<=r) return sums[x];
// go to both left and right side
int m=(lx+rx)/2;
long long s1 = sum(l,r,2*x+1,lx,m);
long long s2 = sum(l,r,2*x+2,m,rx);
return s1+s2;
}

long long sum(int l, int r){
return sum(l,r,0,0,size);
}

};

int main()
{
FAST_INP;
int n,m,v;
cin >> n >> m;
segtree st;
st.init(n);
for(int i=0;i<n;i++){
cin >> v;
st.set(i,v);
}

for(int i=0;i<m;i++){
int op;
cin >> op;
if(op==1) {
int i,v;
cin >> i >> v;
st.set(i,v);
} else {
int l, r;
cin >> l >> r;
cout << st.sum(l,r) << endl;
}
}

return 0;
}



The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

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### Wing-Kam

Consultant by day. Developer by night. AWS certified. Exploring #CloudNative currently.