AtCoder Beginner Contest 050

A

思路：依题意即可。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;
}
int n, k, l;
const int N = 1e6 + 100;
int f1[N];
int f2[N];
int find(int x, int *fa)
{
    return fa[x] == x ? x : fa[x] = find(fa[x], fa);
}
void merge(int x, int y, int *fa)
{
    fa[find(x, fa)] = fa[find(y, fa)];
}
map<pair<int, int>, int> mp;
int main()
{
    int a, b;
    char c;
    cin >> a >> c >> b;
    if (c == '+')
        cout << a + b;
    else
        cout << a - b;
    return 0;
}

B

思路：依题意即可。

记录总和即可。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;
}
int n, k, l;
const int N = 1e6 + 100;
int f1[N];
int f2[N];
int find(int x, int *fa)
{
    return fa[x] == x ? x : fa[x] = find(fa[x], fa);
}
void merge(int x, int y, int *fa)
{
    fa[find(x, fa)] = fa[find(y, fa)];
}
map<pair<int, int>, int> mp;
int main()
{

    int n;
    cin >> n;
    ll sum = 0;
    vector<ll> a(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        sum += a[i];
    }
    int m;
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        int j, k;
        cin >> j >> k;
        cout << sum - a[j] + k << "\n";
    }
    return 0;
}

C

思路：乘法原理+特判。

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
ll quickPow(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}

void init()
{
    // 求阶乘
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
    }
    // 求逆元
    inv[maxn - 1] = quickPow(fac[maxn - 1], mod - 2);
    for (int i = maxn - 2; i >= 0; i--)
    {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
ll C(int n, int m)
{
    if (m > n)
    {
        return 0;
    }
    if (m == 0)
        return 1;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
ll get(ll a, ll b, ll c, ll d)
{
    return C(c - a + d - b, c - a) % mod;
}
int n, k, l;
const int N = 1e6 + 100;
int f1[N];
int f2[N];
int find(int x, int *fa)
{
    return fa[x] == x ? x : fa[x] = find(fa[x], fa);
}
void merge(int x, int y, int *fa)
{
    fa[find(x, fa)] = fa[find(y, fa)];
}
map<pair<int, int>, int> mp;
int main()
{
    int n;
    cin >> n;
    vector<int> a(n + 1);
    map<int, int> mp;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        mp[a[i]]++;
    }
    if (n & 1)
    {
        bool flag = 0;
        if (mp[0] != 1)
        {
            cout << 0 << '\n';
            return 0;
        }
        for (int i = 2; i <= n - 1; i += 2)
        {
            if (mp[i] != 2)
            {
                cout << 0 << '\n';
                return 0;
            }
        }
        cout << quickPow(2, n / 2) << '\n';
    }
    else
    {
        for (int i = 1; i <= n; i += 2)
        {
            if (mp[i] != 2)
            {
                cout << 0 << '\n';
                return 0;
            }
        }
        cout << quickPow(2, n / 2) << '\n';
    }
    return 0;
}

D

若 \((u,v)\) 合法（即存在 \(a,b\)），则 \((2u,2v)\)，\((2u+1,2v+1)\)，\((2u,2v+2)\) 也必定合法。考虑后面新增一位，分讨 \(a,b\) 新增 \((0,0),(1,0),(1,1)\) 可得。

设 \(f(n)\) 表示 \(a+b=v\le n\) 的合法 \((u,v)\) 数。有转移 \(f(n)=f(\dfrac{n}{2})+f(\dfrac{n-1}{2})+f(\dfrac{n-2}{2})\)，除法下取整。

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// LUOGU_RID: 173256890
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
ll inv[maxn], fac[maxn]; // 分别表示逆元和阶乘
// 快速幂
map<ll, ll> f;
ll cal(ll n)
{
    if (f[n])
    {
        return f[n];
    }
    return f[n] = (cal(n / 2) + cal((n - 1) / 2) + cal((n - 2) / 2)) % mod;
}
int main()
{

    ll n;
    cin >> n;
    f[0] = 1;
    f[1] = 2;
    ll ans = cal(n);
    cout << ans << '\n';
    return 0;
}

本次收获：领悟了一种奇妙的二进制递归？