Codeforces Round #341 (Div. 2) ABCDE

http://www.cnblogs.com/wenruo/p/5176375.html

A. Wet Shark and Odd and Even

题意：输入n个数，选择其中任意个数，使和最大且为奇数。

题解：算出所有数的和，如果奇数的个数为奇数个，则减去最小的奇数，否则不用处理。

 #include <bits/stdc++.h>

 using namespace std;

 #define PI acos(-1.0)

 #define EXP exp(1.0)

 #define ESP 1E-6

 #define clr(x,c) memset(x,c,sizeof(x))

 typedef long long ll;

 using namespace std;

 int main()

 {

     int n;

     int cnt = ; int minn = ;

     int val;

     ll ans = ;

     scanf("%d", &n);

     for (int i = ; i < n; ++i) {

         scanf("%d", &val);ans += val;

         if (val %  == ) {

             cnt++;

             minn = min(minn, val);

         }

     }

     if (cnt %  == ) ans -= minn;

     cout << ans;

     return ;

 }

B. Wet Shark and Bishops

题意：一个1000*1000的棋盘，上面摆放n个棋子，同一对角线上的两个棋子可以相互攻击，求一共有多少对棋子可以相互攻击。

题解：有主对角线和副对角线，对于每一个对角线，设对角线上有棋子X个，则该对角线有C(X,2)对棋子可以相互攻击。

#include <bits/stdc++.h>

#define PI acos(-1.0)

#define EXP exp(1.0)

#define ESP 1E-6

#define clr(x,c) memset(x,c,sizeof(x))

typedef long long ll;

using namespace std;

int xSubY[];

int xAddY[];

ll c[][];

//c[i][j] := C(i, j)

void init()

{

    c[][] = ;

    for(int i = ; i <= ; i++)

    {

        c[i][] = ;

        c[i][i] = ;

        for(int j = ; j < i; j++)

            c[i][j] = (c[i-][j-] + c[i-][j]);

    }

}

int main()

{

    init();

    int n;

    int x, y;

    scanf("%d", &n);

    for (int i = ; i < n; ++i) {

        scanf("%d%d", &x, &y);

        xSubY[x-y+]++;

        xAddY[x+y]++;

    }

    ll ans = ;

    for (int i = ; i <= ; ++i) {

        ans += c[ xSubY[i] ][];

        ans += c[ xAddY[i] ][];

    }

    cout << ans;

    return ;

}

C. Wet Shark and Flowers

题意：n个鲨鱼围成一圈1~n，每个鲨鱼有一个数，鲨鱼i的数字在范围l[i]~r[i]，如果相邻数字的乘积为P或P的倍数（p为素数），则两个鲨鱼各获得1000元，求一共能获得钱的期望。

题解：分别考虑相邻的每两个就可以，对于相邻两个数，只有至少一个为p的倍数，才有可能乘积是p的倍数。对于每个鲨鱼一共有r-l+1个数字，有r/p-(l-1)/p的p的倍数。具体看代码。

#include <bits/stdc++.h>

#define PI acos(-1.0)

#define EXP exp(1.0)

#define ESP 1E-6

#define clr(x,c) memset(x,c,sizeof(x))

typedef long long ll;

using namespace std;

const int N = ;

int l[N];

int r[N];

double ans[N];

int main()

{

    int n, p;

    scanf("%d%d", &n, &p);

    for (int i = ; i < n; ++i) {

        scanf("%d%d", &l[i], &r[i]);

    }

    l[n] = l[]; r[n] = r[];

    for (int i = ; i < n; ++i) {

        double a = r[i] - l[i] + ;

        double b = r[i + ] - l[i + ] + ;

        double pa = r[i] / p - (l[i] - ) / p;

        double pb = r[i + ] / p - (l[i + ] - ) / p;

        double tmp = (pa * b + pb * a - pa * pb) / (a * b);

        ans[i] += tmp;

        ans[i + ] += tmp;

    }

    double res = ;

    for (int i = ; i <= n; ++i) res += ans[i];

    printf("%f", res * );

    return ;

}

D. Rat Kwesh and Cheese

题意：很简单，求a₁ = x^{y^z};a₂ = x^{z^y};a₃ = (x^y)^z;a₄ = (x^z)^y;a₅ = y^{x^z};a₆ = y^{z^x};a₇ = (y^x)^z;a₈ = (y^z)^x;a₉ = z^{x^y};a₁₀ = z^{y^x};a₁₁ = (z^x)^y;a₁₂ = (z^y)^x.的最大值。(0.1 ≤ x, y, z ≤ 200.0)

题解：很容易想到求对数，但是，y^z也会很大，所以想当然的求了两次对数，也就是log(log(x^y^z))变成z*log(y)+log(log(x));但是log(x)会出现负数……（后来发现好多人都是这么做的，笑）。正解是取一次对数，然后使用long double，精度足够。

#include <bits/stdc++.h>

typedef long double lld;

using namespace std;

const lld ESP = 1E-;

lld cal(lld x, lld y, lld z)

{

    //return z * log(y) + log(log(x));

    return log(x) * pow(y, z);

}

lld ca(lld x, lld y, lld z)

{

    //return log(y * z * log(x));

    return y * z * log(x);

}

int main()

{

    const char* result[] = {"x^y^z", "x^z^y", "(x^y)^z", "(x^z)^y", "y^x^z",

                        "y^z^x", "(y^x)^z", "(y^z)^x", "z^x^y", "z^y^x", "(z^x)^y", "(z^y)^x"};

    lld a[];

    lld x, y, z;

    cin >> x >> y >> z;

    a[] = cal(x, y, z);

    a[] = cal(x, z, y);

    a[] = ca(x, y, z);

    a[] = ca(x, z, y);

    a[] = cal(y, x, z);

    a[] = cal(y, z, x);

    a[] = ca(y, x, z);

    a[] = ca(y, z, x);

    a[] = cal(z, x, y);

    a[] = cal(z, y, x);

    a[] = ca(z, x, y);

    a[] = ca(z, y, x);

    lld maxn = a[];

    int res = ;

    for (int i = ; i < ; ++i) {

        if (maxn < a[i] - ESP) {

            maxn = a[i];

            res = i;

        }

    }

    printf("%s", result[res]);

    return ;

}

E. Wet Shark and Blocks

题意：n个堆，每个堆有b个数字，每一个堆取一个数字，第i个堆取一个数字做第i位的数字，构成一个n位数，求构成的数字被x除余k有多少种方案。

题解：看到这题就想到了前几天做的题http://www.cnblogs.com/wenruo/p/5153790.html 不过这个简单些，然并卵，比赛时并没有写出来……dp+矩阵快速幂

#include <bits/stdc++.h>

typedef long long ll;

using namespace std;

int digit[];

const ll MOD = ;

typedef vector<ll> vec;

typedef vector<vec> mat;

mat mul(mat &A, mat &B)

{

    mat C(A.size(), vec(B[].size()));

    for (unsigned int i = ; i < A.size(); ++i) {

        for (unsigned int k = ; k < B.size(); ++k) {

            for (unsigned int j = ; j < B[].size(); ++j) {

                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;

            }

        }

    }

    return C;

}

mat pow(mat A, int n)

{

    mat B(A.size(), vec(A.size()));

    for (unsigned int i = ; i < A.size(); ++i)

        B[i][i] = ;

    while (n > ) {

        if (n & ) B = mul(B, A);

        A = mul(A, A);

        n >>= ;

    }

    return B;

}

/*

dp[i][k] i位数 余数为k

dp[i+1][(j+k*10)%x] = dp[i][k]*digit[j], 1<=j<=9

*/

ll cal(int n, int x, int k)

{

    mat f(x, vec(x));

    for (int i = ; i < x; ++i) {           // 上一位余数

        for (int j = ; j <= ; ++j) {      // 首位数字

            int num = (i *  + j) % x;     // 当前余数

            f[num][i] += digit[j];

        }

    }

    mat v(x, vec());

    for (int i = ; i <= ; ++i)

        v[i % x][] += digit[i];    // dp[1][i%x][i]

    f = pow(f, n - );

    v = mul(f, v);

    return v[k][];

}

int main()

{

    int n, b, k, x;

    scanf("%d%d%d%d", &n, &b, &k, &x);

    int val;

    for (int i = ; i < n; ++i) {

        scanf("%d", &val);

        digit[val]++;

    }

    cout << cal(b, x, k);

    return ;

}

总体而言这次的比赛比较简单^_^