题解报告（CDUT暑期集训——第六场）
题解报告（CDUT暑期集训——第六场）
A - oval-and-rectangle
HDU - 6362
-
思路：水题 积分一化就出来了
-
AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
#include<string>
#include<sstream>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll Pow_mod(ll a, ll b, ll c){
ll ans = 1;
a %= c;
while (b){
if (b & 1) ans = (ans * a) % c;
a = (a * a) % c;
b >>= 1;
}
return (ans % c);
}
int gcd(int a, int b){
return b ? gcd(b, a % b) : a;
}
int lcm(int a, int b){
return a * 1ll / gcd(a, b) * b;
}
const double pi = acos(-1);
const double eps = 5 * 1e-7;
int t;
double a, b, ans;
int main(){
scanf("%d", &t);
while (t -- ){
scanf("%lf%lf", &a, &b);
ans = 2 * b + a * pi;
printf("%.6lf\n", ans - eps);
}
return 0;
}
B - bookshelf
HDU - 6363
-
__思路：根据斐波拉契数列和gcd的各种性质 得到ans与gcd的关系 然后对约数使用容斥原理 还要用欧拉降幂防止数据溢出 __
-
AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
#include<string>
#include<sstream>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll Pow_mod(ll a, ll b, ll c){
ll ans = 1;
a %= c;
while (b){
if (b & 1) ans = (ans * a) % c;
a = (a * a) % c;
b >>= 1;
}
return (ans % c);
}
int gcd(int a, int b){
return b ? gcd(b, a % b) : a;
}
int lcm(int a, int b){
return a * 1ll / gcd(a, b) * b;
}
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
int tot;
int primes[N], mu[N];
bool vis[N];
void Mobius(){
memset(vis, false, sizeof(vis));
mu[1] = 1;
tot = 0;
for (int i = 2; i < N; i ++ ){
if (!vis[i]){
primes[tot ++] = i;
mu[i] = -1;
}
for (int j = 0; j < tot && i * primes[j] < N; j ++ ){
vis[i * primes[j]] = true;
if (i % primes[j] == 0){
mu[i * primes[j]] = 0;
break;
}
else mu[i * primes[j]] = -mu[i];
}
}
}
ll fib[N], fact[N << 1], inv[N << 1], inv_f[N << 1];
ll get_inv(ll n){
return Pow_mod(n, mod - 2, mod);
}
void init(){
Mobius();
fib[0] = 0, fib[1] = 1;
for (int i = 2; i < N; i ++ )
fib[i] = (fib[i - 1] + fib[i - 2]) % (mod - 1);
fact[0] = 1; // ÀÛ³Ë
for (int i = 1; i < N << 1; i ++ )
fact[i] = (i * fact[i - 1]) % mod;
inv[1] = 1; // ÄæÔª
for (int i = 2; i < N << 1; i ++ )
inv[i] = ((mod - mod / i) * inv[mod % i]) % mod;
inv_f[0] = 1; // ÀÛ³ËÄæÔª
for (int i = 1; i < N << 1; i ++ )
inv_f[i] = (inv_f[i - 1] * inv[i]) % mod;
}
ll C(ll n, ll m){
return fact[n] * inv_f[m] % mod * inv_f[n - m] % mod;
}
int t, n, k;
ll ans, tmp;
int main(){
init();
scanf("%d", &t);
while (t -- ){
scanf("%d%d", &n, &k);
ans = 0;
for (int i = 1; i <= n; i ++ ){
if (n % i) continue;
tmp = 0;
for (int j = i; j <= n; j += i){
if (n % j) continue;
tmp = (tmp + 1ll * mu[j / i] * C(n / j + k - 1, k - 1) % mod + mod) % mod;
}
ans = (ans + tmp * (Pow_mod(2ll, fib[i], mod) - 1 + mod) % mod + mod) % mod;
}
tmp = C(n + k - 1, k - 1);
ans = ans * Pow_mod(tmp, mod - 2, mod) % mod;
printf("%lld\n", ans);
}
return 0;
}
L - Pinball
HDU - 6373
-
思路：水题 图一画就出来了（最开始想错了样例都过不去卡了半小时
-
AC代码
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
#include<string>
#include<sstream>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll Pow_mod(ll a, ll b, ll c){
ll ans = 1;
a %= c;
while (b){
if (b & 1) ans = (ans * a) % c;
a = (a * a) % c;
b >>= 1;
}
return (ans % c);
}
int gcd(int a, int b){
return b ? gcd(b, a % b) : a;
}
int lcm(int a, int b){
return a * 1ll / gcd(a, b) * b;
}
const double g = 9.8;
const double pi = acos(-1);
int t;
double a, b, x, y, ans, alpha, t1, t2, s1, s2, gx, gy, h;
int main(){
scanf("%d", &t);
while (t -- ){
scanf("%lf%lf%lf%lf", &a, &b, &x, &y);
x = -x;
alpha = atan2(b, a);
h = y - x * tan(alpha);
s1 = x / cos(alpha) + h * sin(alpha);
s2 = h * cos(alpha);
gx = g * sin(alpha);
gy = g * cos(alpha);
t1 = sqrt(2 * s1 / gx);
t2 = sqrt(2 * s2 / gy);
ans = (t1 - t2) / (2 * t2);
printf("%d\n", (int)ans + 1);
}
return 0;
}
