第一周 7.10-7.16

假装自己复活辣。

7.10

CF 689 D Friends and Subsequences

二分。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 10;
int n, a[maxn], b[maxn];

// RMQ
int M[maxn][20], m[maxn][20];
void init()
{
    for(int i = 1; i <= n; i++) M[i][0] = a[i];
    for(int j = 1; (1 << j) <= n; j++)
        for(int i = 1; i + ( 1 << j ) - 1 <= n; i++)
            M[i][j] = max(M[i][j-1] , M[i+(1<<j-1)][j-1]);

    for(int i = 1; i <= n; i++) m[i][0] = b[i];
    for(int j = 1; (1 << j) <= n; j++)
        for(int i = 1; i + ( 1 << j ) - 1 <= n; i++)
            m[i][j] = min(m[i][j-1] , m[i+(1<<j-1)][j-1]);
}

int M_query(int l, int r)
{
    int k = 0;
    while( ( 1 << (k + 1) ) <= r - l + 1 ) k++;
    return max(M[l][k], M[r-(1<<k)+1][k]);
}

int m_query(int l, int r)
{
    int k = 0;
    while( ( 1 << (k + 1) ) <= r - l + 1 ) k++;
    return min(m[l][k], m[r-(1<<k)+1][k]);
}

int main(void)
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", a + i);
    for(int i = 1; i <= n; i++) scanf("%d", b + i);
    init();

    LL ans = 0LL;
    for(int l = 1; l <= n; l++)
    {
        int t1, t2;
        int L = l, R = n, mid;

        if(M_query(l, n) < m_query(l, n)) R = n + 1;
        else while(L < R)
        {
            mid = L + (R - L) / 2;
            if(M_query(l, mid) < m_query(l, mid)) L = mid + 1;
            else R = mid;
        }
        t1 = R;

        L = l, R = n;
        if(M_query(l, n) - 1 < m_query(l, n)) R = n + 1;
        else while(L < R)
        {
            mid = L + (R - L) / 2;
            if(M_query(l, mid) - 1 < m_query(l, mid)) L = mid + 1;
            else R = mid;
        }
        t2 = R;

        ans = ans + t2 - t1;
    }

    printf("%I64d
", ans);

    return 0;
}

7.12

HDU 5514 Frogs

因为每只步长为a[i]的青WA能踩到所有gcd(a[i], m)的倍数石头。

问题转化为给n个数，求能被这n个数中至少一个整除的数之和。

能被一个数整除的所有数之和是一个等差数列 容易计算，但是被多个数整除的数会重， 所以容斥一下。

因为所有gcd都是m的因子，先枚举m的所有因子，容易证明1e9内因子数目不会很多的。

比如 2^2 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19 = 58198140，也就3^2 * 2^6 = 576个因子。

容斥的过程是 求 sigma能被一个gcd整除的数 - sigma能被两个gcd整除的数 +sigma能被三个gcd整除的数 ……

对于每个因子 算它的贡献的时候 它的所有倍数的贡献都被算过了 所以给倍数减去贡献就好了 负贡献也一样。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e4 + 10;
int a[maxn], f[maxn], c[maxn];

int gcd(int a, int b)
{
    return a % b ? gcd(b, a % b) : b;
}

int main(void)
{
    int T;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        int n, m;
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d", a + i);

        memset(f, 0, sizeof(f));
        for(int i = 1; i * i <= m; i++)
        {
            if(m % i == 0)
            {
                f[++f[0]] = i;
                if(i * i != m) f[++f[0]] = m / i;
            }
        }
        sort(f + 1, f + 1 + f[0]);

        memset(c, 0, sizeof(c));
        for(int i = 1; i <= n; i++)
        {
            int x = gcd(a[i], m);
            for(int j = 1; j <= f[0]; j++)
                if(f[j] % x == 0) c[j] = 1;
        }

        LL ans = 0LL;
        for(int i = 1; i < f[0]; i++)
        {
            ans = ans + (LL) (m - 1) / f[i] * ((m - 1) / f[i] + 1) * f[i] / 2 * c[i];
            for(int j = i + 1; j <= f[0]; j++)
                if(f[j] % f[i] == 0) c[j] -= c[i];
        }

        printf("Case #%d: %I64d
", kase, ans);
    }
    return 0;
}

7.13

HDU 5515 Game of Flying Circus

嘻嘻嘻

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int main(void)
{
    int t;
    scanf("%d", &t);
    for(int kase = 1; kase <= t; kase++)
    {
        double T, V1, V2;
        scanf("%lf %lf %lf", &T, &V1, &V2);
        printf("Case #%d: ", kase);
        if(V1 == 0) puts("No");
        else if(V1 >= V2) puts("Yes");
        else if(V1 * sqrt(2) > V2)
        {
            double tmp = V1 / V2 * V1 / V2;
            double a = tmp - 1, b = 600 * tmp, c = a * 90000;
            double x = (- b + sqrt(b * b - 4 * a * c)) / 2 / a;
            if((x + 600) / V1 < T + (600 - x) / V2) puts("Yes");
            else puts("No");
        }
        else if(V1 * 3 > V2)
        {
            double tmp = V2 / V1 * V2 / V1;
            double a = tmp - 1, b = 1800, c = tmp * 90000 - 810000;
            double x = (- b + sqrt(b * b - 4 * a * c)) / 2 / a;
            if((sqrt((300 - x) * (300 - x) + 90000) + 900) / V1 < T + (300 + x) / V2) puts("Yes");
            else puts("No");
        }
        else puts("No");
    }
    return 0;
}

HDU 5478 Can you find it

嘿嘿嘿

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;

LL qpow(LL a, LL b, LL mod)
{
    LL ret = 1LL;
    while(b)
    {
        if(b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}

int main(void)
{
    int kase = 0;
    int C, k1, b1, k2;
    while(~scanf("%d %d %d %d", &C, &k1, &b1, &k2))
    {
        printf("Case #%d:
", ++kase);
        int ok = 0;
        for(int a = 1; a < C; a++)
        {
            int b = C - qpow(a, k1 + b1, C);
            if(qpow(a, k1, C) == qpow(b, k2, C))
            {
                ok = 1;
                printf("%d %d
", a, b);
            }
        }
        if(!ok) puts("-1");
    }
    return 0;
}

7.14

HDU 5534 Partial Tree

高老师太神拉。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int f[2222], dp[2222];

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        for(int i = 1; i < n; i++) scanf("%d", f + i);
        for(int i = 2; i < n; i++)
        {
            dp[i] = f[i] + (i - 2) * f[1];
            for(int j = 2; j <= i - j + 1; j++)
                dp[i] = max(dp[i], dp[j] + dp[i-j+1]);
        }
        printf("%d
", dp[n-1] + 2 * f[1]);
    }
    return 0;
}

HDU 5492 Find a path

The magic values are no greater than 30.

#include <iostream>
#include <cstdio>
using namespace std;
int G[31][31], dp[31][31][1801];

int main(void)
{
    int T;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        int N, M;
        scanf("%d %d", &N, &M);
        for(int i = 1; i <= N; i++)
            for(int j = 1; j <= M; j++)
                scanf("%d", &G[i][j]);

        for(int i = 0; i <= N; i++)
            for(int j = 0; j <= M; j++)
                for(int k = 0; k <= 1800; k++)
                    dp[i][j][k] = 1e9;

        dp[1][1][G[1][1]] = 0;
        for(int i = 1; i <= N; i++)
        {
            for(int j = 1; j <= M; j++)
            {
                if(i == 1 && j == 1) continue;
                for(int k = 0; k <= 1800; k++)
                {
                    if(dp[i-1][j][k] != 1e9) dp[i][j][k+G[i][j]] = min(dp[i][j][k+G[i][j]], ((dp[i-1][j][k] + k * k) / (i + j - 2) + G[i][j] * G[i][j]) * (i + j - 1) - (k + G[i][j]) * (k + G[i][j]));
                    if(dp[i][j-1][k] != 1e9) dp[i][j][k+G[i][j]] = min(dp[i][j][k+G[i][j]], ((dp[i][j-1][k] + k * k) / (i + j - 2) + G[i][j] * G[i][j]) * (i + j - 1) - (k + G[i][j]) * (k + G[i][j]));
                }
            }
        }

        int ans = 1e9;
        for(int i = 0; i <= 1800; i++) ans = min(ans, dp[N][M][i]);
        printf("Case #%d: %d
", kase, ans);

    }
    return 0;
}