Codeforces Round #594 (Div. 2)
Codeforces Round #594 (Div. 2)
A. Integer Points
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题意：给出若干个\(y = x + p\) 和 \(y = -x + q\) 求它们交点坐标为整数的个数
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思路：\(y = x + p\) 与 \(y = -x + q\) 的交点为 \(\left(\frac{q - p}{2}, \frac{p + q}{2}\right)\) 即只用统计p和q数组中的奇偶数即可
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AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e5 + 10;
int t, n, m;
ll cnt1, cnt2, cnt3, cnt4, ans;
ll p[N], q[N];
int main(){
// freopen("my_in.txt", "r", stdin);
// freopen("my_out.txt", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
ans = 0, cnt1 = 0, cnt2 = 0, cnt3 = 0, cnt4 = 0;
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> p[i];
if (p[i] & 1)
cnt1 ++ ;
else
cnt2 ++ ;
}
cin >> m;
for (int i = 1; i <= m; i ++ ){
cin >> q[i];
if (q[i] & 1)
cnt3 ++ ;
else
cnt4 ++ ;
}
ans = cnt1 * cnt3 + cnt2 * cnt4;
cout << ans << "\n";
}
return 0;
}
B. Grow The Tree
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题意：给出若干长度的线段 构成一个折线 要求折线的起点到终点的最长距离
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思路：先排序 排序后以前一半作为所求线段在\(x\)轴的投影 以后一半作为所求线段在\(y\)轴的投影 这样求出来的线段长度是最大的
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AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e5 + 10;
int n;
int a[N];
ll x, y, ans;
int main(){
// freopen("my_in.txt", "r", stdin);
// freopen("my_out.txt", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i = 1; i <= n; i ++ )
cin >> a[i];
sort(a + 1, a + n + 1);
for (int i = 1; i <= n / 2; i ++ )
x += a[i];
for (int i = n / 2 + 1; i <= n; i ++ )
y += a[i];
ans = x * x + y * y;
cout << ans << "\n";
return 0;
}
C. Ivan the Fool and the Probability Theory
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题意：给出一个\(n * m\)的方格 要在尽可能多的小格子中填入东西 每个小格子最多能有一个相邻的格子被填充 问一共有多少种填法
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思路：
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AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int n, m;
ll ans;
ll f[N];
void init(){
f[1] = 1, f[2] = 2;
for (int i = 3; i < N; i ++ )
f[i] = (f[i - 1] + f[i - 2]) % mod;
}
int main(){
// freopen("my_in.txt", "r", stdin);
// freopen("my_out.txt", "w", stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
init();
cin >> n >> m;
ans = mult_mod(2, ((f[n] + f[m]) % mod - 1 + mod) % mod, mod);
cout << ans << "\n";
return 0;
}
