Codeforces Round #294 (Div. 2)

 

水 A. A and B and Chess

/*
    水题
*/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
using namespace std;

const int maxn = 1e6 + 10;
int a[maxn];

int main(void)
{
    //freopen ("A.in", "r", stdin);

    string s1;
    int suma = 0;    int sumb = 0;
    for (int i=1; i<=8; ++i)
    {
        cin >> s1;
        for (int j=0; s1[j]!='\0'; ++j)
        {
            if (s1[j] == '.')    continue;
            else if (s1[j] == 'Q')    suma += 9;
            else if (s1[j] == 'R')    suma += 5;
            else if (s1[j] == 'B')    suma += 3;
            else if (s1[j] == 'N')    suma += 3;
            else if (s1[j] == 'P')    suma += 1;
            else if (s1[j] == 'q')    sumb += 9;
            else if (s1[j] == 'r')    sumb += 5;
            else if (s1[j] == 'b')    sumb += 3;
            else if (s1[j] == 'n')    sumb += 3;
            else if (s1[j] == 'p')    sumb += 1;
        }
    }

    if (suma > sumb)    cout << "White" << endl;
    else if (suma < sumb)    cout << "Black" << endl;
    else    cout << "Draw" << endl;

    return 0;
}

水 B. A and B and Compilation Errors

题意：三组数列，依次少一个，找出少了的两个数

思路：

1. 三次排序，逐个对比（如果没找到，那个数在上一个数列的末尾）
2. 求和做差，最简单！

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

const int maxn = 1e5 + 10;
int a[maxn];
int b[maxn];
int c[maxn];

int main(void)
{
    //freopen ("B.in", "r", stdin);

    int n, x, y;

    while (~scanf ("%d", &n))
    {
        x = y = 0;
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &a[i]);
        }
        sort (a+1, a+1+n);
        for (int i=1; i<=n-1; ++i)
        {
            scanf ("%d", &b[i]);
        }
        sort (b+1, b+1+n-1);
        for (int i=1; i<=n-1; ++i)
        {
            if (a[i] == b[i])    continue;
            else
            {
                x = a[i];
                break;
            }
        }
        if (x == 0)        x = a[n];
        for (int i=1; i<=n-2; ++i)
        {
            scanf ("%d", &c[i]);
        }
        sort (c+1, c+1+n-2);
        for (int i=1; i<=n-2; ++i)
        {
            if (b[i] == c[i])    continue;
            else
            {
                y = b[i];
                break;
            }
        }
        if (y == 0)        y = b[n-1];
        printf ("%d\n%d\n", x, y);

    }

    return 0;
}

/*
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 1e5 + 10;
int a[maxn];
int b[maxn];
int c[maxn];
int suma, sumb, sumc;

int main(void)
{
    //freopen ("B.in", "r", stdin);

    int n;

    while (~scanf ("%d", &n))
    {
        suma = sumb = sumc = 0;
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &a[i]);    suma += a[i];
        }
        for (int i=1; i<=n-1; ++i)
        {
            scanf ("%d", &b[i]);    sumb += b[i];
        }
        for (int i=1; i<=n-2; ++i)
        {
            scanf ("%d", &c[i]);    sumc += c[i];
        }

        printf ("%d\n%d\n", suma - sumb, sumb - sumc);
    }

    return 0;
}
*/

构造 C. A and B and Team Training

题意：方案：高手1和菜鸟2 或者 高手2菜鸟1 三人组队求最大组队数
思路：

1. 高手加菜鸟每三个分开，在n，m的数字之内
2. 高手多，高手2；菜鸟多，菜鸟2 比较好理解

#include <cstdio>
#include <algorithm>
using namespace std;

int main(void)
{
    //freopen ("C.in", "r", stdin);

    int n, m;

    while (~scanf ("%d%d", &n, &m))
    {
        int ans = (n + m) / 3;
        ans = min (ans, n);
        ans = min (ans, m);
        printf ("%d\n", ans);
    }

    return 0;
}

/*
#include <cstdio>
#include <algorithm>
using namespace std;

int main(void)
{
    //freopen ("C.in", "r", stdin);

    int n, m;

    while (~scanf ("%d%d", &n, &m))
    {
        int cnt = 0;
        while (n && m && (n + m) >= 3)
        {
            if (n >= m)
            {
                n -= 2;        m -= 1;
            }
            else
            {
                n -=1;        m -= 2;
            }
            cnt++;
        }

        printf ("%d\n", cnt);

    return 0;
}
*/