### Abracadabra

I’ve always be longing to own a blog of my own, but I am never able to deal with the annoying hosting stuff. And today, I finally decided to use wordpress – use money to fill the emptiness of knowledge. That’s fair.

Anyway, I am having this very own blog now. I’m using it to record my coding training for programming contests. I had been using github but… it’s just not the same and you have no idea how I envied others for their blogs.

Hopefully I’ll truly “knee the way to my chosen dream”, and I hope you enjoy reading my tiny steps. ### 2019.5.14 Walkabout    ### 2017 PacNW E Straight Shot

/* 11/16/2017
* Solution: Use binary search to find the appropriate angle.
*/

```#include
#include
#include
#define PI acos(-1)
using namespace std;

int main() {
long long n, x;
double vv;
int ll, rr;
double v;
double summ = 0;
cin >> n >> x >> v;
for (int i = 1; i > ll >> rr >> vv;
summ += (rr - ll) * 1.0 * vv / v;
}
if (summ > 0) {
summ = - summ;
}
double l = 0, r = 90;
double result = 100;
while (abs(result) > 1e-5) {
if (l > 60) {
cout  60) {
cout << "Too hard";
} else {
cout << fixed << setprecision(3) << (double)x / (v * cos(l * PI / 180));
}
return 0;
}
```

### CF 474D Flowers

/* 7/14/2018
* Problem: http://codeforces.com/problemset/problem/474/D
* Solution: Dp. dp[i] = dp[i-1] + dp[i-k]. One state can be transferred to from two sources: 1. eat only white flower from the i-k day 2. otherwise.
* Notes: 1. Remember to mod the MOD every time there’s a summing process.
2. The result of a mod can be negative.
* A little bit gibber-gabber:
For both the notes, I KNEW it but I now don’t..??
What is the point of everything..
*/

```#include
using namespace std;

int main() {
int t, k;
cin >> t >> k;
long long ind = {0}, sum = {0};
int MOD = 1000000007;
for (int i = 0; i < k; i++) {
ind[i] = 1;
}
for (int i = k; i <= 100005; i++) {
ind[i] = (ind[i - 1] + ind[i - k]) % MOD;
}
sum = 1;
for (int i = 0; i  a >> b;
long long ans = (sum[b] - sum[a - 1]) % MOD;
if (ans < 0) {
ans += MOD;
}
cout << ans << endl;
}
return 0;
}

```

### 2017 PacNW L Delayed work

/* 11/14/2017
*/

#include
#include
#include
using namespace std;

int main() {
long long k, p, x;
cin >> k >> p >> x;
int m = ceil(sqrt(k * p / x));
double ans = (double)k * p / m + x * m ;
cout << fixed << setprecision(3) << ans;
return 0;
}

### 2017 PacNW C Fear factoring

/* 11/12/2017
*/

```
#include
#include
using namespace std;

int main() {
long long a,b;
cin >> a >> b;
long long tmp = sqrt(b);
long long ans = 0;
for (long long i = 1; i <= tmp; i++) {
long long lb = max(tmp, (a-1) / i) + 1;
long long ub = max(tmp, b / i);
ans += i * (b / i - (a-1)/ i) + (lb + ub) * (ub - lb + 1) / 2;
}
cout << ans;
return 0;
}
```

### 2017 PacNW E Straight Shot

/* 11/16/2017
* Solution: Use binary search to find the appropriate angle.
*/

```#include
#include
#include
#define PI acos(-1)
using namespace std;

int main() {
long long n, x;
double vv;
int ll, rr;
double v;
double summ = 0;
cin >> n >> x >> v;
for (int i = 1; i > ll >> rr >> vv;
summ += (rr - ll) * 1.0 * vv / v;
}
if (summ > 0) {
summ = - summ;
}
double l = 0, r = 90;
double result = 100;
while (abs(result) > 1e-5) {
if (l > 60) {
cout  60) {
cout << "Too hard";
} else {
cout << fixed << setprecision(3) << (double)x / (v * cos(l * PI / 180));
}
return 0;
}
```

### 2017 PacNW A Odd Palindrome

/* 11/14/2017
* /

```
#include
#include
using namespace std;

int main() {
string s;
cin >> s;
int len = s.length() - 1;
for (int i = 0; i < len; i++) {
if (s[i] == s[i + 1]) {
cout << "Or not." << endl;
return 0;
}
}
cout << "Odd." << endl;
return 0;
}

```