学步园

A: 模拟26进制加法即可

/*************************************************************************
    > File Name: cf-293-a.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月25日 星期三 13时40分05秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

string A, B;

int main ()
{

    while (cin >> A >> B)
    {
        int len = A.length();
        int c = 1;
        for (int i = len - 1; i >= 0; --i)
        {
            if (!c)
            {
                break;
            }
            if (A[i] == 'z' && c)
            {
                A[i] = 'a';
            }
            else if (c)
            {
                ++A[i];
                c = 0;
            }
        }
        if (A >= B)
        {
            cout << "No such string" << endl;
            continue;
        }
        cout << A << endl;
    }
    return 0;
}

B:注意读题仔细,另外要先计算第一个答案,标记好,再去算第二个答案

/*************************************************************************
    > File Name: cf-293-b.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月25日 星期三 01时01分17秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 200100;
char s[N], t[N];
bool vis[N];
int num[130];

int main ()
{
    while (~scanf("%s%s", s, t))
    {
        int ans1 = 0, ans2 = 0;
        int len = strlen (t);
        memset (num, 0, sizeof(num));
        memset (vis, 0, sizeof(vis));
        for (int i = 0; i < len; ++i)
        {
            ++num[t[i]];
        }
        len = strlen (s);
        for (int i = 0; i < len; ++i)
        {
            if (num[s[i]] > 0)
            {
                vis[i] = 1;
                ++ans1;
                --num[s[i]];
            }
        }
        for (int i = 0; i < len; ++i)
        {
            if (!vis[i])
            {
                if (s[i] >= 'A' && s[i] <= 'Z')
                {
                    if (num[s[i] - 'A' + 'a'] > 0)
                    {
                        ++ans2;
                        --num[s[i] - 'A' + 'a'];
                    }
                }
                else
                {
                    if (num[s[i] - 'a' + 'A'] > 0)
                    {
                        ++ans2;
                        --num[s[i] - 'a' + 'A'];
                    }
                }
            }
        }
        printf("%d %d\n", ans1, ans2);
    }
    return 0;
}

C:水题模拟题

/*************************************************************************
    > File Name: cf-293-c.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月25日 星期三 14时36分58秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 100100;
vector <int> scr[N];
int arr[N];
int at[N];
int ord[N];

int main ()
{
    int n, m, k;
    int app;
    while (~scanf("%d%d%d", &n, &m, &k))
    {
        for (int i = 1; i <= n; ++i)
        {
            scr[i].clear();
        }
        for (int i = 1; i <= n; ++i)
        {
            int pos = i / k + ((i % k) ? 1 : 0);
            scanf("%d", &app);
            at[app] = pos;
            scr[pos].push_back (app);
            if (i % k)
            {
                ord[app] = i % k - 1;
            }
            else
            {
                ord[app] = k - 1;
            }
        }
        LL ans = 0;
        for (int i = 1; i <= m; ++i)
        {
            scanf("%d", &app);
            ans += at[app];
            int s = at[app];
            int cur = ord[app];
            if (s == 1 && cur == 0)
            {
                continue;
            }
            if (cur == 0)
            {
                int size = scr[s - 1].size();
                int &a = scr[s - 1][size - 1];
                int &b = scr[s][cur];
                at[a] = s;
                at[b] = s - 1;
                ord[a] = 0;
                ord[b] = size - 1;
                swap (a, b);
            }
            else
            {
                int &a = scr[s][cur - 1];
                int &b = scr[s][cur];
                ord[a] = cur;
                ord[b] = cur - 1;
                swap (a, b);
            }
        }
        printf("%I64d\n", ans);

}

D:水题概率dp
设dp[i][j] i秒时,车厢里j个人的概率

/*************************************************************************
    > File Name: cf-293-d.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月25日 星期三 01时59分30秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

double dp[2010][2010];
int n, t;
double p;

int main ()
{
    while (~scanf("%d%lf%d", &n, &p, &t))
    {
        memset (dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for (int i = 1; i <= t; ++i)
        {
            dp[i][0] = dp[i - 1][0] * (1 - p);
            int e = min (n, i);
            for (int j = 1; j <= e; ++j)
            {
                if (j < e)
                {
                    dp[i][j] = dp[i - 1][j - 1] * p + dp[i - 1][j] * (1 - p);
                }
                else
                {
                    dp[i][j] = dp[i - 1][j - 1] * p + dp[i - 1][j];
                }
            }
        }
        double ans = 0;
        for (int i = 0; i <= t; ++i)
        {
            ans += dp[t][i] * i;
        }
        printf("%.10f\n", ans);
    }
    return 0;
}