AtCoder Beginner Contest 172 部分题解

Posted 2020-12-13 hznumqf

C：给两个栈，每次只能取栈顶元素，取完后自动pop  问能取到最多几个元素

栈中元素之和必须小于等于K

官方题解给出的做法是O（N+M） 受上一场CF启发，此题可以很自然联想到二分做法。  二分答案，答案显然具有单调性。check函数只需遍历一遍可能情况

复杂度O((N+M)logX) 

int n, m;
ll k;
ll a[maxn], b[maxn];
ll sum1[maxn];
ll sum2[maxn];

bool check(int x) {
    for (int i = x - m ; i <= min(x,n); i++) {
        if (sum1[i] + sum2[x - i] <= k) {
            //cout << x << " " << i <<" "<< x - i << endl;
            return 1;
        }
    }
    return 0;
}

int main() {
    scanf("%d%d%lld", &n, &m, &k);
    for (int i = 1; i <= n; i++) {
        scanf("%lld", a + i );  sum1[i] = sum1[i - 1] + a[i];
    }
    for (int i = 1; i <= m; i++) {
        scanf("%lld", b + i); sum2[i] = sum2[i - 1] + b[i];
    }
    int l = 0, r = n + m;
    while (l < r) {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    if (check(l)) printf("%d", l);
    else printf("%d", l-1);
}

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D:

函数F(x) 表示x的因子个数 。 求 sgma i*F(i)          N<=1e7

利用d(i)==(每个质因数的指数+1)的积 跑一遍素数筛顺便可筛出F函数，之后相乘即可

int n;
ll d[maxn], minn[maxn], prime[maxn];
bool is[maxn];
int tot;

void init() {
    d[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (is[i] == 0) prime[++tot] = i, d[i] = 2, minn[i] = 1;
        for (re int j = 1; prime[j] * i <= n && j <= tot; j++)
        {
            is[i * prime[j]] = 1;
            if (i % prime[j] == 0)
            {
                minn[i * prime[j]] = minn[i] + 1;
                d[i * prime[j]] = d[i] / (minn[i] + 1) * (minn[prime[j] * i] + 1);
                break;
            }
            minn[i * prime[j]] = 1;
            d[i * prime[j]] = d[i] * 2;
        }
    }
}

int main() {
    scanf("%d", &n);
    init();
    ll res = 0;
    for (int i = 1; i <= n; i++) {
        res += i * d[i];
    }
    printf("%lld", res);
}

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E：

构造一对数组a，b  要求满足 数组中数字均不超过M。 且 1.a中元素互不相同 2.b中元素互不相同 3.ai不等于bi  问可构造出多少对

1<=N<=M<=5e5

 

推公式即可  先把a数组定下来，有A(n,m) 种选法。 再考虑所有情况减去不对的情况。 故容斥（实际上是错排的推广） 

注意最后不对的情况要除以（m-n）！去重

注意取逆元。

ll inv[maxn];
ll n, m;

void inver(ll p) {
    inv[0] = 1;
    inv[1] = 1;
    for (int i = 2; i <= m; i++) {
        inv[i] = (p - p / i) * inv[p % i] % p;
    }
}


ll c[maxn];
ll fac[maxn];

ll quickPower(ll a, ll b, ll m) {   //计算a的b次方
    ll ans = 1;
    ll base = a;
    while (b) {
        if (b & 1) {
            ans *= base;
            ans %= m;
        }
        base *= base;
        base %= m;
        b >>= 1;   //注意是b>>=1 not b>>1
    }
    return ans;
}

ll get_inv(ll x) {
    return quickPower(x, MOD - 2, MOD);
}

void get_fac() {
    fac[0] = 1;
    for (int i = 1; i <= m; i++) {
        fac[i] = (fac[i - 1] * i)%MOD;
    }
}

void get_C(ll n) {
    c[0] = 1;
    for (int i = 0; i < n; i++) c[i + 1] = ((c[i] * (n - i)) % MOD * inv[i + 1]) % MOD;
}

int main() {
    scanf("%lld%lld", &n, &m);
    inver(MOD);
    get_fac();
    get_C(n);
    ll res = (1 * ((fac[m] * get_inv(fac[m - n])) % MOD * get_inv(fac[m - n])) % MOD)%MOD;
    ll tmp = fac[m];
    int f = 1;
    for (int i = 1; i <= n; i++) {
        tmp = (tmp + ((-1ll * f) * ((c[i] * fac[m - i]) % MOD))+MOD) % MOD;
        f *= -1;
    }
    //cout << res << tmp << endl;
    printf("%lld", (tmp * res) % MOD);
}

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F：

NIm博弈  不同之处在于 后手者可以在游戏开始前取出第一堆中的若干个放到第二堆中 问最少放几个可以保证他必胜 若不能输出-1

2<=N<=300     1<=ai<=1e12

ll a[305];
ll m, s;

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%lld", a + i);
    m = a[2];
    for (int i = 3; i < n; i++) m ^= a[i];
    s = a[0] + a[1];
    ll rem = s - m;
    if ((rem % 2) || (rem < 0)) {
        puts("-1");
        return 0;
    }
    rem /= 2;
    ll b = rem;
    int f = 1;
    for (ll i = 50; i >= 0; i--) {
        ll p = 1ll << i;
        if ((m & p) && (rem & p)) f = 0;
        if ((m & p) && ((b + p) <= a[0])) b += p;
    }
    b = a[0] - b;
    if ((b >= a[0]) || (b < 0)) f = 0;
    if (f) printf("%lld", b);
    else puts("-1");
}

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